Answer:
Yes
Explanation:
How many atoms are in 9.35 moles of lithium?
Answer:
Answer: 3.45 moles Li contains 2.08 × 10 (to the power of)24 atoms . Explanation: The relationship between atoms and moles is:. 1 mole atom. =6.022
Answer:
\(\boxed {\sf 5.63 \times 10^{24} \ atoms \ Li}}\)
Explanation:
We are asked to find how many atoms are in 9.35 moles of lithium (Li)
Moles are converted to atoms using Avogadro's Number or 6.022×10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of lithium.
We convert using dimensional analysis, so we set up a conversion factor using Avogadro's Number.
\(\frac {6.022 \times 10^{23} \ atoms \ Li}{1 \ mol \ Li}\)
We are converting 9.35 moles of lithium to atoms, so we multiply the conversion factor by this value.
\(9.35 \ mol \ Li *\frac {6.022 \times 10^{23} \ atoms \ Li}{1 \ mol \ Li}\)
The units of moles of lithium cancel.
\(9.35 *\frac {6.022 \times 10^{23} \ atoms \ Li}{1 }\)
\(9.35 *{6.022 \times 10^{23} \ atoms \ Li}\)
\(5.63057\times10^{24} \ atoms \ Li\)
The original value of moles (9.35) has 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 0 in the thousandth place tells us to leave the 3 in the hundredth place.
\(5.63 \times 10^[24} \ atoms \ Li\)
9.35 moles of lithium contains approximately 5.63×10²⁴ atoms of lithium.
A sample of neon initially has a volume of 2.80 L at 23 degrees Celsius. What is final temperature, in degrees Celsius, is needed to change the volume of the gas to .75 L? Please show your work in order to receive credit.
The relation between the volume and the temperature of the gas is given by Charles's law. The final temperature of the gas at 0.75 liters is -193.8°C.
What is Charles's law?Charles's law was derived from the ideal gas equation and is used to state the relationship between the temperature and the volume of the gas. With a decrease in volume the temperature decreases.
If the pressure is kept constant then with an increase in temperature the volume of the gas expands. The law is given as,
V₁ ÷ T₁ = V₂ ÷ T₂
Given,
Initial volume (V₁) = 2.80 L
Initial temperature (T₁) = 23 °C = 296.15 K
Final volume (V₂) = 0.75 L
Final temperature = T₂
Substituting the values above as:
T₂ = (V₂ × T₁) ÷ V₁
= 0.75 × 296.15 ÷ 2.80
= 79.325 K
Kelvin is converted as, 79.325K − 273.15 = -193.8°C
Therefore, the final temperature is -193.8°C.
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Sharing a custom report will share the report configuration and data included in the report.
Sharing a custom report will share the report configuration and data included in the report: False.
What is a custom report?In Go-ogle analytics, a custom report can be defined as a report which is created by an end user and it's based on the following:
DimensionHow it is shared.MetricsGenerally, sharing a custom report would not share the report configuration and data included in the report because only the configuration information are shared while user data remains private.
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Which of the following would have the strongest ion-dipole interaction with water? Explain.
T13+
Cd²+
Na+
The strongest ion-dipole interaction with water is T13+ than Cd²+ than Na+.
What is ion dipole interaction?An ion-dipole interaction is defined as the intermolecular force of attraction present between the charged ion (anion or cation) and the molecule. It is mainly found in the solution where the ionic compounds dissolved in the polar solvents.
The strength of ion-dipole interactions is mainly dependent on the distance and the charge. Ionic charge goes on increasing moving across a period on the periodic table.
Since, charge on Ti is 3 and greater than Cd which is greater than Na.
Thus, we concluded that the strongest ion-dipole interaction with water is T13+ than Cd²+ than Na+.
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Take a balloon that we filled up with our breath. Tell me what would happen if we put it outside and let it sit in the sun for a bit. Support your answer using the KMT.Keep in mind we are going to be under the impression that no air can escape from the balloon.
The balloon will be filled by a gas, and as the gas is not escaping we can assume that it is at a constant pressure.
The KMT (Kinetic Molecular Theory) explains the macroscopic properties of gases. This theory states that gases are formed by particles in constant motion, and that their kinetic energy is directly proportional to the temperature of the gas.
If we leave the balloon filled with gas in the sun we assume that the temperature of the gas inside will start to rise. When the temperature starst to rise, and the pressure is constant, particles will gain kinetic energy. This will result in particles moving faster and coliding with each other and with the balloon's walls more often. As a consecuence the particles will tend to stay farther from one another and the volume of the balloon will increase.
This is an example of Charles's Law that states that at constant pressure the volume of a gas is directly proportional to its temperature. In other words, when the temperature of the gas increases the volume increases as well.
To summarize, if we fill a balloon and we leave it in the sun its volume will increase.
Based on a Kc value of 0.250 and the given data table, what are the equilibrium concentrations of XY, X, and Y , respectively?
From the solution that we have in the question;
The concentration of X and Y is 0.28 MThe concentration of XY is 0.32 MWhat is the equilibrium constant?The equilibrium constant, denoted as K, is a value that quantitatively represents the ratio of the concentrations of products to reactants at equilibrium in a chemical reaction.
It is a fundamental concept in chemical equilibrium.
The value of the equilibrium constant provides valuable information about the position of equilibrium and the relative concentrations of species involved in a chemical reaction.
Kc = [X] [Y]/[XY]
\(0.25 = (0.1 + x)^2/(0.5 - x)\)
\(0.25(0.5 - x) = (0.1 + x)^2\)
\(0.125 - 0.25x =0.01 + 0.2x + x^2\\ x^2 + 0.45x - 0.115 = 0\)
x = 0.18 M
The equilibrium amount of X and Y= 0.28 M and the equilibrium concentration of XY = 0.32 M
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Based on the answer to the question that we have;
A 0.28 M concentration of X and Y exists at equilibriumXY's concentration at equilibrium is 0.32 M.The equilibrium constantThe ratio of the product to reactant concentrations in a chemical reaction at equilibrium is represented quantitatively by the equilibrium constant, abbreviated as K.
It is a cornerstone of the theory of chemical equilibrium.
A chemical reaction's equilibrium position and the relative concentrations of the species involved can both be learned from the equilibrium constant's value.
Kc = [X][Y]/[XY]
\(0.25 = (0.1 + x)^2/(0.5 - x)\\0.25(0.5 - x) = (0.1 +x)^2\\0.125 - 0.25x = 0.01 +0.2x +x^2\\= 0.18 M\)
The equilibrium concentration of;
XY =0.5 - 0.18
=0.32 M
Then the equilibrium amount of
X and Y is
0.1 + 0.18= 0.28 M.
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How does the chemical structure of a substance affect its interaction with other substances?
This is due to the fact that a substance's chemical qualities, such as its molecular form, polarity, and functional groups, govern how it behaves and interacts with other substances.
How does their chemical makeup impact their chemical characteristics?By illustrating the spatial arrangement of atoms and chemical bonds within the molecule, chemical structure establishes the molecular geometry of a compound. In doing so, chemists are given a crucial visual depiction of a chemical formula.
In what ways do drugs interact with one another?In a chemical reaction, reactants come into contact with one another, atoms in the reactants break their connections with one another, and then the atoms reorganise and form new bonds to create the products.
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Consider the equilibrium X ⥫⥬ Y, where the forward reaction rate constant is greater than the reverse reaction rate constant. Which of the following is true about the equilibrium constant: (a) Kc > 1, (b) Kc < 1, (c) Kc = 1?
Answer:
the correct answer is (a) Kc > 1.
The ratio of the concentrations of the products to the reactants, with each concentration increased to the power of its stoichiometric coefficient, determines the equilibrium constant (Kc) in a chemical reaction.
What is forward reaction ?It is implied that the forward reaction is preferred and moves along more quickly than the reverse reaction in the case presented when the forward reaction rate constant is larger than the reverse reaction rate constant. This implies that the concentration of the products relative to the reactants will be higher at equilibrium.
Consequently, the following is the accurate statement on the equilibrium constant (Kc): (a) Kc > 1
Since the ratio of product concentrations to reactant concentrations, which determines Kc, is favoured in this situation and the products, the equilibrium constant will be greater than 1.
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Convert 57.0 degrees C todegrees F.[?] °FEnterT in degrees F
Answer
134.6 °F
Explanation
To covert degrees C to degrees F use:
(0°C × 9/5) + 32 = 32°F
Therefore 0 °C = 32 °F
We are given 57.0 °C
(57.0°C x 9/5) + 32 = 134.6 °F
Chemistry Lab Determination of the Universal Gas Constant (R)
SHOW ALL WORK
Given:
Initial mass of butane lighter: 54.24g
Final Mass of Butane Lighter: 54.01g
Temperature of water: 23.0°C
Volume of gas collected: 100.0mL
FIND:
Barometric pressure of room: 766.86 mmHg CONVERTED TO atm
Vapor pressure of water at room temperature(PH2O) (IN atm)
FIND:
Mass difference if butane lighter in grams
Moles of Butane gas collected in moles of C4H10
Partial pressure if butane gas in atm
Converted temperature of water in Kelvin
Converted volume of gas collected in Liters
Experimental value of R in Latm/molk
Accepted value of R in Latm/molk
Percent error in experimental value of R in %
CONCLUSION QUESTIONS:
1. List at least 3 factors that either did it could contribute to the percent error
2. Should the value of R go up or down if the gas had not been corrected for the partial pressure of water. Why?
3. How could this experiment be repeated to increase the accuracy, or in other words, decrease the percent error?
NOTE: LET ME KNOW IF YOU WANT A PICTURE OF THE LAB INSTRUCTIONS TO HELP SOLVE
ALSO SHOW ALL WORK PLS
To determine the universal gas constant (R), we can use the ideal gas law equation: PV = nRT, where P is the total pressure, V is the volume of gas, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.
Barometric pressure of the room: To find the barometric pressure in atm, we convert 766.86 mmHg to atm by dividing it by 760 mmHg/atm, giving us 1.0089 atm.Vapor pressure of water at room temperature (PH2O).The vapor pressure of water at 23.0°C is 0.0313 atm.
Mass difference of butane lighter:The mass difference is calculated by subtracting the initial mass (54.24 g) from the final mass (54.01 g), resulting in a difference of 0.23 g.
Moles of butane gas collected:To find the moles of butane gas, we can use the equation n = m/M, where n is the number of moles, m is the mass, and M is the molar mass of butane (58.12 g/mol). Thus, n = 0.23 g / 58.12 g/mol = 0.003959 mol.
Partial pressure of butane gas:The partial pressure of butane gas is calculated by multiplying the moles of butane gas (0.003959 mol) by the ideal gas constant (R) and the converted temperature (23.0°C + 273.15 K). Let's assume the converted temperature is 296.15 K.
Converted volume of gas collected:The volume of gas collected is given as 100.0 mL, which needs to be converted to liters by dividing it by 1000, resulting in 0.1 L.
Experimental value of R:The experimental value of R can be determined by rearranging the ideal gas law equation to solve for R: R = (P - PH2O) * V / (n * T).
Accepted value of R:The accepted value of R is 0.0821 Latm/molK.
The percent error can be calculated using the formula: (|Experimental value - Accepted value| / Accepted value) * 100.
Factors contributing to percent error could include experimental error in mass measurements, inaccurate temperature measurements, and loss of gas during collection or transfer.
If the gas had not been corrected for the partial pressure of water, the value of R would be lower because the partial pressure of water would contribute to the total pressure, resulting in a smaller value for P in the ideal gas law equation.
To increase accuracy and decrease percent error, the experiment could be repeated multiple times to obtain an average value, use more precise measuring instruments, conduct the experiment in a controlled environment, and ensure accurate calibration of equipment.
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The correct IUPAC name for the structure shown is
A)
ethylmethylamine.
B)
methylamine.
C)
ethylamine.
D)
ethylmethylhydridoamine.
Answer:
A
Explanation:
it has a methyl group, ethyl group and amine group
The correct IUPAC name for the structure shown in the provided image is "ethylamine." The structure consists of a central nitrogen atom bonded to two carbon atoms.
The correct IUPAC name for the structure shown in the provided image is "ethylamine." The structure consists of a central nitrogen atom bonded to two carbon atoms. According to the IUPAC naming rules, the longest carbon chain is selected as the parent chain, which in this case consists of two carbon atoms. The substituent attached to the parent chain is an ethyl group, denoted as "C2H5". The amine functional group, which consists of the nitrogen atom, is named as "amine". Since there is only one amine group attached to the carbon chain, it is referred to as "ethylamine." Therefore, option C) "ethylamine" is the correct IUPAC name for the given structure.
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A continous fractionating column separates 1500 kg/h of a solution of benzene and toluene containing 0.75 mass benzene into an overhead product containing 0.925 mass fraction benzene and bottom product containing 0.075 mass fraction of benzene. A reflux ration of 2.5 kg of reflux per kg of product is to be used. Calculate the quanity of top and bottom product in kg/h.
The quantity of the top product is approximately 1483.4 kg/h, and the quantity of the bottom product is approximately 1116.7 kg/h.
What is the quantity of top and bottom products in kg/h?We solve using the concept of material balances.
Given:
F = Feed rate of the solution (benzene + toluene) = 1500 kg/h
Fb = Feed rate of benzene in the feed = 0.75 * F = 1125 kg/h
Ft = Feed rate of toluene in the feed = F - Fb = 375 kg/h
Benzene Balance:
Fb = Bt + Bb
where Bt is the flow rate of benzene in the top product and Bb is the flow rate of benzene in the bottom product.
1125 kg/h = Bt + Bb
Toluene Balance:
The mass balance equation for toluene can be written as:
Ft = Tt + Tb
where Tt is the flow rate of toluene in the top product and Tb is the flow rate of toluene in the bottom product.
375 kg/h = Tt + Tb
The reflux ratio is given as 2.5 kg reflux per kg of product
The total product flow rate (Pt) will be:
Pt = Bt + Tt = Bb + Tb
Therefore, the reflux flow rate (R) is:
R = 2.5 * Pt
Solving the equations, we find:
Bt = 1116.7 kg/h
Tb = 8.3 kg/h
Top Product Flow Rate = Bt + Tt = Bt + (Ft - Tb) = Bt + (375 - 8.3) = 1116.7 + 366.7 = 1483.4 kg/h
Bottom Product Flow Rate = Bb + Tb = 1125 - 8.3 = 1116.7 kg/h
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Strontium hydroxide reacts with hydrobromic acid to produce Strontium bromide and
water.
Write and balance the chemical reaction above, use it for problems 1-4 below:
1. If 5.50 moles of strontium hydroxide were consumed, how much moles of water are
produced?
2. Find the mass of hydrobromic acid used to produce 7.50 moles water.
3. If 10.8 g of strontium hydroxide were used, how much moles of strontium bromide are
produced?
4. If 13.3 g of hydrobromic acid were consumed, find the mass of the water produced.
Please need this ASAP. Calculate the mass of lime, CaO, that would be produced from 250 tonnes of limestone,
CaCO3.
Relative atomic masses: C 12; O 16; Ca 40.
Answer:
1.4×10⁸ g of CaO
Explanation:
We'll begin by converting 250 tonnes to grams (g). This can be obtained as follow:
1 tonne = 1×10⁶ g
Therefore,
250 tonne = 250 × 1×10⁶
250 tonne = 2.5×10⁸ g
Next, the balanced equation for the reaction.
CaCO₃ —> CaO + CO₂
Next, we shall determine the mass of CaCO₃ that decomposed and the mass CaO produced from the balanced equation. This can be obtained as follow:
Molar mass of CaCO₃ = 40 + 12 + (16×3)
= 40 + 12 + 48
= 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
Molar mass of CaO = 40 + 16
= 56 g/mol
Mass of CaO from the balanced equation = 1 × 56 = 56 g
SUMMARY:
From the balanced equation above,
100 g of CaCO₃ decomposed to produce 56 g of CaO.
Finally, we shall determine the mass of CaO produced by the decomposition of 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃. This can be obtained as follow:
From the balanced equation above,
100 g of CaCO₃ decomposed to produce 56 g of CaO.
Therefore, 2.5×10⁸ g of CaCO₃ will decompose to produce =
(2.5×10⁸ × 56)/100 = 1.4×10⁸ g of CaO.
Thus, 1.4×10⁸ g of CaO will be obtained from 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃.
would you be able to help me?
Answer:
yes...
Explanation:
wha
How many grams of CuSO4 must be dissolved in 575.9 ml. of total solution to create a 1.6114 M solution of CuSO4?O a. 148.1O b. 5.96e-4O c.0.7264O d. 38.85
Answer:
a. 148.1 (g).
Explanation:
What is given?
Volume of solution = 575.9 mL = 0.5759 L,
Molarity of solution = 1.6114 M,
Molar mass of CuSO4 = 159.5 g/mol.
Step-by-step solution:
To find the mass of CuSO4 in the solution, we have to use the molarity formula:
\(Molarity\text{ \lparen M\rparen=}\frac{moles\text{ of solute}}{liters\text{ of solution}}=\frac{mol}{L}.\)Let's solve for 'moles of solute', and replace the given data (remember that the volume must be in liters), like this:
\(\begin{gathered} Moles\text{ of solute=molarity}\cdot liters\text{ of solution,} \\ \\ Moles\text{ of solute=1.6114 M}\cdot0.5759\text{ L,} \\ \\ Moles\text{ of solute}=0.928\text{ moles.} \end{gathered}\)The final step is to convert 0.928 moles of CuSO4 to grams using the molar mass of CuSO4, as follows:
\(0.928\text{ moles CuSO}_4\cdot\frac{159.5\text{ g CuSO}_4}{1\text{ mol CuSO}_4}=148.02\text{ g.}\)The answer would be the nearest to 148.02 g, which is a. 148.1 (g).
Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-13 at 25°C)
Answer: when concentrations of acid and base are same, pH = pKa
PH = 12.38 pOH = 1.62
Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00
The pH of the solution is 12.38 and the pOH of the solution is 1.62.
Using the Henderson Hasselbalch equation;
pH = pKa + log [A-]/[HA]
Where;
pKa = - log Ka = -log[ 4.2x10-13] = 12.38
[A-] = [PO43–] = 0.285 M
[HA] = [HPO42-] = 0.285 M
Substituting values;
pH = 12.38 + [ 0.285 M]/[ 0.285 M]
pH = 12.38
But;
pH + pOH = 14
pOH = 14 - 12.38
pOH = 1.62
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True or False. Water pressure increases as you travel deeper.
Answer:
The answer is true.
Explanation:
The deeper you go the more water there's on top of you pushing down.
Answer:
True
Explanation:
Fresh vegetables with high water content do not freeze well. Explain.
a) water expands whenit freezes and damagesthecell wall.
b) water contracts when it freezes and damagesthe cell wall.
c) water expands whenit freezes and does not damagethecell wall.
d) water contracts when it freezes and does not damagethecell wall.
e) water does not change in size whenit freezes.
Answer:
a) water expands when it freezes and damages the cell wall.
Explanation:
We know that most liquids expand when heated and contract when cooled. Water does not behave in the same way.
Water contracts on cooling from any temperature until 4°C when it begins to expand until 0°C. Hence a given mass of water has its least volume and highest density at 4°C.
Hence, fresh vegetables with high water content do not freeze well because water expands when it freezes and damages the cell wall.
Nitrates are _____.
compounds made from carbon dioxide
used by plants to make protein
released into the atmosphere by respiration
are harmful to the environment
Answer:
used by plants to make protein
Explanation:
so B
Calculate the vapor pressure (in torr) at 298 K in a solution prepared by dissolving 46.8 g of the non-volatile non-electrolye glucose in 117 g of methanol. The vapor pressure of methanol at 298 K is 122.7 torr. Enter your answer to 2 decimal places.
Answer: The total partial pressure of the solution is 131.37 torr.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) ......(1)
For glucose:Given mass of glucose = 46.8 g
Molar mass of glucose = 180 g/mol
Plugging values in equation 1:
\(\text{Moles of glucose}=\frac{46.8g}{180g/mol}=0.26 mol\)
For methanol:Given mass of methanol = 117 g
Molar mass of methanol = 32 g/mol
Plugging values in equation 1:
\(\text{Moles of methanol}=\frac{117g}{32g/mol}=3.66 mol\)
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
\(\chi_A=\frac{n_A}{n_A+n_B}\) .....(2)
where n is the number of moles
Putting values in equation 2:
\(\chi_{methanol}=\frac{3.66}{0.26+3.66}=0.934\)
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture. The equation for Raoult's law follows:
\(p_A=\chi_A\times p_T\) .....(3)
where \(p_A\) is the partial pressure of component A in the mixture and \(p_T\) is the total partial pressure of the mixture
We are given:
\(p_{methanol}=122.7torr\\\chi_{methanol}=0.934\)
Putting values in equation 3, we get:
\(122.7torr=0.066\times p_T\\\\p_T=\frac{122.7torr}{0.934}=131.37torr\)
Hence, the total partial pressure of the solution is 131.37 torr.
HQ5.40
Homework Answered Due Today, 11:59 PM
The reaction 3H₂(g) + N₂(g) → 2NH3(g) has an enthalpy of reaction of -92.6 kJ/mol. If 1 g of hydrogen and 2 g of nitrogen are
reacted, how much heat is produced (kJ)?
The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ
How do i determine the heat energy produced?First, we shall obtain the limiting reactant. Details below:
3H₂ + N₂ -> 2NH₃
Molar mass of N₂ = 28 g/molMass of N₂ from the balanced equation = 1 × 28 = 28 g Molar mass of H₂ = 2 g/molMass of H₂ from the balanced equation = 3 × 2 = 6 gFrom the balanced equation above,
28 g of N₂ reacted with 6 g of H₂
Therefore,
2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂
We can see that only 0.43 g of H₂ is needed in the reaction.
Thus, the limiting reactant is N₂
Finally, we the amount of heat energy produced. Details below:
3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ
Molar mass of N₂ = 28 g/molMass of N₂ from the balanced equation = 1 × 28 = 28 gFrom the balanced equation above,
When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.
Therefore,
When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ
Thus the heat energy produced from the reaction is -6.61 KJ
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Balance the equations by putting the necessary coefficients in the blanks. Normally we do not write 1s when balancing, but for this particular question you need to include them for full credit. __Na3N___ Na +__ N2 ___H3PO4 + __ KOH __K3PO4 + __ H2O __ N2 +__ H2 __ NH3 __H2O2 __ O2 + __ H2O __ Zn + __ HCl __ ZnCl2 + __H2 __ C2H6 + __ O2 __ CO2 + __H2O __ CuCl2 + __H2S __ CuS + __HCl
Balancing a chemical equation is the process of ensuring that the number of atoms of each element in the reactants is equal to the number of atoms of that same element in the products.
Balance the chemical eqations given in the problem?
Na3N → 3 Na + ½ N2H3PO4 + 3 KOH → K3PO4 + 3 H2ON2 + 3 H2 → 2 NH3H2O2 → O2 + 2 H2OZn + 2 HCl → ZnCl2 + H2C2H6 + 7/2 O2 → 2 CO2 + 3 H2OCuCl2 + H2S → CuS + 2 HClChemical equations are used to describe the reactants and products in a chemical reaction. These equations are written using chemical formulas and symbols, indicating the types and numbers of atoms or molecules involved in the reaction. However, these equations must be balanced to obey the law of conservation of mass, which states that the total mass of the reactants must equal the total mass.
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The mass of a carbon atom in grams is approximately 12 amu, or 2.0 × 10-23 g.
To make things easy we could ask, how many carbon atoms would reach an easy-to-remember mass, 12 g? (12 g is easy to remember because it's already on the periodic table for carbon!)
Let's do it! Calculate the number of carbon atoms we would need to reach 12 g on a lab balance.
It may be helpful to consider the following equation to solve:
(mass of carbon atom in grams) × N = 12 g of carbon (reading on balance)
Where N is the number of atoms needed to reach 12 g of carbon on the lab's balance. (check within this question for the other information you need)
How many carbon atoms, N, are needed to reach 12 grams on the scale?
Therefore, we need 6.0 × 1022 atoms of carbon to reach 12 g on the lab balance.
What is carbon ?Carbon is an element found in nature that is essential for life on Earth. It is an abundant, non-metallic, and highly versatile element that is used in a variety of ways in many different fields. Carbon is found in all living organisms and is the basis of all life on Earth. It is also found in rocks, soil, and the atmosphere. Carbon is the fourth most abundant element in the universe and is the basis for organic compounds that form the basis of all living things. In industry, carbon is used to produce steel, glass, plastics, and many other materials. Carbon is also used in the manufacture of fuels and is a component of the chemical processes necessary for life. Carbon is also used in batteries, semiconductors, and solar cells.
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800 j of heat absorbed by 120 g of water at 25 c, what is the final temperature at equilibrium
The final temperature at equilibrium will be approximately 26.525 °C.
How to determine the final temperatureTo find the final temperature at equilibrium, we can use the equation:
Q = m * c * ΔT
Where:
Q = heat energy absorbed or released
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
In this case, 800 J of heat is absorbed by 120 g of water at 25 °C. We need to find the final temperature at equilibrium.
Using the equation for water:
Q_water = m_water * c_water * ΔT_water
Rearranging the equation, we can solve for the change in temperature (ΔT_water):
ΔT_water = Q_water / (m_water * c_water)
Substituting the given values:
ΔT_water = 800 J / (120 g * 4.18 J/g°C)
Calculating:
ΔT_water ≈ 1.525 °C
To find the final temperature at equilibrium, we add the change in temperature to the initial temperature:
Final temperature = 25 °C + 1.525 °C
Final temperature ≈ 26.525 °C
Therefore, the final temperature at equilibrium will be approximately 26.525 °C.
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4. Compare the conductivity observations of dilute and concentrated NiCl₂ and NaOH.
Explain why you observed this.
NiCl₂ and NaOH are both electrolytes, which means that they dissociate in water to form ions, which are charged particles that can conduct electricity.
When NiCl₂ is dissolved in water, it dissociates into Ni²⁺ ions and Cl⁻ ions. The conductivity of a dilute solution of NiCl₂ is relatively low because there are fewer ions present in the solution, and thus fewer ions available to conduct electricity.
When the concentration of NiCl₂ is increased, the conductivity of the solution also increases. This is because as the concentration of the solution increases, there are more ions present in the solution, which results in a greater ability to conduct electricity.
Similarly, when NaOH is dissolved in water, it dissociates into Na⁺ ions and OH⁻ ions. As with NiCl₂, the conductivity of a dilute solution of NaOH is relatively low because there are fewer ions present in the solution. As the concentration of NaOH is increased, the conductivity of the solution increases because there are more ions present in the solution.
In summary, the conductivity of dilute solutions of NiCl₂ and NaOH is low because there are fewer ions present in the solution. As the concentration of these solutions increases, the conductivity increases because there are more ions present in the solution.
how many moles are in 150 grams lf NaOH
Answer:
3.75 mole
Explanation:
molecular mass of NaOH = 39.97 gm
i.e,
==>39.97 gm NaOH contains 1 mole of NaOH
==>1 gm of NaOH contains 1/39.97 mole of NaOH
==>150 gm of NaOH contains (1/39.97) *150 mole of NaOH
==>150 gm of NaOH contains 3.75 mole of NaOH.
b) Ammonia and sulfuric acid react according to the equation given below. How many millilitres of 0.110 M sulfuric acid are required to neutralize exactly 25.0 mL of 0.0840 M NH3 solution? 2 NH3(aq) + H₂SO4 (aq) → (NH4)2SO4(aq)
The amount of 0.110 M sulfuric acid are required to neutralize exactly 25.0 mL of 0.0840 M NH3 solution is 9.55mL.
A Neutralization Reaction: What Is It?A neutralisation reaction is a chemical process in which an acid and a base combine to produce salt and water as the end products. H+ ions and OH- ions combine to generate water during a neutralisation process.
2 NH3(aq) + H₂SO4 (aq) → (NH4)2SO4(aq)
moles of NH3 = (25ml x 1L/1000mL) x 0.084M
=> 2.1x 10^(-3) moles
The mole ratio of NH3 to H₂SO4 in the given reaction
=> moles of H₂SO4 = 2.1 x 10^(-3) moles of NH3 x 1 molesH₂SO4/2 moles
NH3
=> 1.05 x 10^(-3) moles
Volume = moles/molarity
=> 1.05 10^(-3) moles/0.110M
=> 9.55 x 10^(-3) L = 9.55mL
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What is it called when a substance changes?
Chemical elements of chemical compounds help compensate substances, which are made up of one or more reactants (also known as reactants) and one or more products (also known as products).
What is the process of a material changing shape called?Chemical synthesis, or, alternatively, chemical breakdown into two or more separate molecules, occurs when one material reacts with another to create a new substance.These processes are referred to as chemical reactions, and they are typically irreversible barring additional chemical reactions.
What else do you call a chemical change?An alteration of one or more compounds as one or more unique and fresh substances is referred to as a chemical change or chemical reaction.In those other terms, a chemical is an atomic rearrangement-based chemical process.
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the relative formula masses (Mr) are: CaCo3 = 100; CaO =56 ; Co2=44
describe how this experiment could be used to provide evidence for the law of conservation of mass.
[6 marks]
include your answer:
-method
-which measurements should eb taken
-how the student could show evidence for the conservation for mass
The law of conservation of mass states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. To provide evidence for this law, we can perform an experiment in which calcium carbonate (\(CaCO_3\)) is decomposed to produce calcium oxide (CaO) and carbon dioxide (\(CO_2\) ), and then measure the masses of the reactants and products.
Method:
Weigh a sample of \(CaCO_3\) using a balance.
Heat the \(CaCO_3\) in a crucible until it decomposes to CaO and \(CO_2\). The \(CO_2\) gas will escape, leaving only CaO in the crucible.
Allow the crucible to cool and then weigh it again to determine the mass of the CaO produced.
Collect the \(CO_2\) gas that is released during the reaction in a gas syringe or other collection device. Measure the volume of \(CO_2\) gas produced, and calculate its mass using its molecular weight.
Which measurements should be taken:
The following measurements should be taken:
The mass of the \(CaCO_3\) used as a reactant.
The mass of the CaO produced as a product.
The volume of \(CO_2\) gas produced during the reaction.
The temperature and pressure of the \(CO_2\) gas to allow for the calculation of its mass.
How the student could show evidence for the conservation of mass:
To show evidence for the law of conservation of mass, the student can compare the mass of the \(CaCO_3\) used as a reactant to the total mass of the products, which includes the mass of CaO produced and the mass of \(CO_2\) gas released.
The sum of the masses of CaO and \(CO_2\) should be equal to the mass of the \(CaCO_3\) used as a reactant, within experimental error. This will provide evidence that the mass of the reactants is conserved and equals the mass of the products, as required by the law of conservation of mass.
Additionally, the student could calculate the theoretical yield of CaO and CO2 based on the balanced equation for the reaction, and compare this to the actual yield obtained from the experiment. Any difference between the theoretical and actual yields could be due to experimental error, but the comparison can still provide additional evidence for the conservation of mass.
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