a chemistry student collected 0.032l of h2gas at 1.1 atm pressure and 24 C using the following chemical reaction how man grams of magnesium must have reacted.

Mg(s) + HCL (aq) to H2 9g) + MgCL2 (aq)

Answer:

0.0104 moles of gas in the flask.

Explanation:

To calculate the number of moles of gas in the flask, you can use the ideal gas law equation: PV = nRT. Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is temperature.

First, you need to convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin. The pressure in atm is 760.0 mmHg / 760 mmHg/atm = 1 atm. The temperature in Kelvin is 17.00°C + 273.15 = 290.15 K.

Next, you need to convert the volume from mL to L. The volume in L is 250.0 mL / 1000 mL/L = 0.2500 L.

Now you can plug all the values into the ideal gas law equation and solve for n: (1 atm)(0.2500 L) = n(0.08206 L·atm/mol·K)(290.15 K). Solving for n gives n = 0.0104 mol.

So there are approximately 0.0104 moles of gas in the flask.

Answer:

1.96 × 10²⁴ molecules

Explanation:

Step 1: Given data

Mass of ethanol (m): 150. g

Step 2: Calculate the number of moles (n) corresponding to 150. g of ethanol

The molar mass of ethanol is 46.07 g/mol.

150. g × 1 mol/46.07 g = 3.26 mol

Step 3: Calculate the number of molecules in 3.26 moles of ethanol

To convert moles into molecules, we need Avogadro's number: there are 6.02 × 10²³ molecules in 1  mole of molecules.

3.26 mol × 6.02 × 10²³ molecules/1  mol = 1.96 × 10²⁴ molecules

Answer:

: kinetic energy is the energy transformation that occurs in a hot balloon.

Explanation:

Hot air balloons use a propane burner that converts chemical energy to thermal energy. The hot air is less dense than than the colder air and it lifts the balloon

Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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Answer:

f a runner decreases his velocity from +20 m/s to +10 m/s, with average acceleration is -5m/s².

Explanation:

Answer:

A

Refer to pic for explanation

Based on the given reaction, the acid-base pairs in this reaction are:

An acid-base pair refers to a set of two chemical species that are related through the transfer of a proton (H+ ion) during a chemical reaction.

One species acts as an acid by donating a proton, while the other acts as a base by accepting that proton.

In the given reaction:

HCO₃⁻ (aq) + NH₃ (aq) → NH₄⁺ + CO₃²⁻

An acid-base pair can be identified as follows:

HCO₃⁻ (bicarbonate ion) can act as an acid by donating a proton (H⁺), becoming CO₃⁻.

NH₃ (ammonia) can act as a base by accepting a proton (H⁺), becoming NH₄⁺ (ammonium ion).

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Answer:

Explanation:

The atomic weight of iron (Fe) is 55.845 u. Since it's impossible to count the amount of atoms in a given mass of a substance, we measure atoms in moles. No matter what the element or compound is, there will be \(6.022 * 10^{23}\) atoms in one mole of compound. This is called Avogadro's number. Atomic weight, also called molar mass, is measured in grams per mole. We first find the number of moles of iron in 3.47 grams by setting up an inequality: \(\frac{55.845 g}{1 mol} = \frac{3.47 g}{x mol}\). Solving this inequality shows that there are 0.06213627 moles of iron in 3.47 grams. Now, all we do is multiply that value by Avogadro's number, \(6.022*10^{23}\). This gives us 3.7418 * 10^22 atoms in 3.47 grams of iron.

Answer:

I believe it is the Ethiopian ibex

Explanation:

Answer:

I think it the Ethiopian ibex

Explanation:

im srry if it wrong.

hope this helps tho

The volume of avocado here is 250 ml as per the given data i.e., mass of 215 g and a density of 0.86 g/mL.

The density of an artifact is delineated as its mass divided by the volume. Density is commonly expressed in grams per cubic centimeter.

The grams are a unit of mass and indeed cubic centimeters are a unit of volume. A box with more particles will be denser than a box with fewer particles.

The density of a solid, liquid, or gas illustrates how closely packed the particles can be. The amount of mass per unit volume is defined as density.

Density is defined as d = M/V, where d is density, M is mass, and V is volume. Density is commonly measured in grams per cubic centimetre.

We know that,

Density = mass/volume.

Volume = mass/density.

Here, it is given that

Density = 0.86 g/mL.

Mass = 215 g.

Volume = 215/0.86 mL.

Volume = 250 mL.

Thus, the answer is 250 mL.

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Benzene

Low denisty object than the water usually floats on the water.

Another example: orange peel which floats on water has density of about 0.84 g/cm³

(a) Condensation reaction is a reaction whereby two molecules combine to form a single molecule. During this reaction, water is removed. For example, two amino acids combine by a covalent bond, water molecule is then removed as a second product. Esterification is a type of condensation reaction, whereby an ester is formed from an alcohol and a carboxylic acid. Another example of condensation reaction is saponification which describes the alkaline hydrolysis reaction of an ester.

(b) Addition reaction is a reaction whereby a double bond is broken by additing an element. Unsaturated organic molecules participate in addition reactions, this is because unsaturated organic molecules are those that have double bonds.

For example addition of hydrogen: Addition of hydrogen to a carbon-carbon double bond is called hydrogenation, which results in the removal of the double bond.

Explanation:

this explanation may help u to understand:)

The energy found in the reactants remains in the system, and the reactants also take energy from the surroundings.

Answer:nonmetal

Explanation: gggg

Be(OH)₂ compound is a strong electrolyte.

A solid or liquid that totally or almost completely ionises or dissociates in a solution is considered a strong electrolyte. In the solution, these ions function well as electrical conductors. An early definition of a "strong electrolyte" was a substance that is an effective conductor of electricity when dissolved in water. Strong electrolytes are those that totally dissociate or ionise when placed in aqueous solution. These electrolytes have a high electrical conductivity and a greater ionisation extension. NaCl, HCl, NaOH, etc. are a few examples.

Correct option: Be(OH)₂

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Answer:

.217, .311, and .472, respectively.

Explanation:

The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).

X of helium=3.50/16.10 = .217

X of krypton=5.00/16.10 = .311

X of neon=7.60/16.10 = .472

Answer:

Chemists generally use the mole as the unit for the number of atoms or molecules of a material. One mole (abbreviated mol) is equal to 6.022×1023 molecular entities (Avogadro's number), and each element has a different molar mass depending on the weight of 6.022×1023 of its atoms (1 mole).\

Have a great day!!!!

Explanation:

1 mole of (NH₄)₂CO₃ has:

Let's consider the chemical formula of ammonium carbonate: (NH₄)₂CO₃.

The subscripts inform us of the number of moles of atoms in 1 mole of molecules.

Nitrogen has 2 subscripts: "1" (which is not written) and "2" (outside the parenthesis). The global subscript for nitrogen is obtained by multiplying both subscripts.

\(N: 1 \times 2 = 2\)

Hydrogen has 2 subscripts: "4" and "2" (outside the parenthesis). The global subscript for nitrogen is obtained by multiplying both subscripts.

\(H: 4 \times 2 = 8\)

The subscript for carbon is "1".

The subscript for oxygen is "3".

1 mole of (NH₄)₂CO₃ has:

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The 3.13 g of K would be needed to completely react with the remaining \(Cl_2\).

To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.

The balanced chemical equation for the reaction between solid potassium and chlorine gas is:

2 K(s) + \(Cl_2\)(g) -> 2 KCl(s)

From the equation, we can see that 2 moles of K react with 1 mole of \(Cl_2\) to form 2 moles of KCl.

First, we need to convert the masses of K and \(Cl_2\) into moles:

moles of K = 15.20 g / 39.10 g/mol = 0.388 mol

moles of \(Cl_2\) = 2.830 g / 70.90 g/mol = 0.040 mol

Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:

Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl

Theoretical yield of KCl from \(Cl_2\): 0.040 mol \(Cl_2\) x (2 mol KCl / 1 mol \(Cl_2\)) = 0.080 mol KCl

We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from \(Cl_2\) is 0.080 mol. Therefore, the limiting reactant is \(Cl_2\), since it produces less product.

To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.

We know that 1 mole of \(Cl_2\) reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining \(Cl_2\) can be calculated as follows:

moles of K needed = 0.040 mol \(Cl_2\) x (2 mol K / 1 mol \(Cl_2\))

                                = 0.080 mol K

This means that 0.080 moles of K would be needed to completely consume the remaining \(Cl_2\). We can convert this to a mass by multiplying by the molar mass of K:

mass of K needed = 0.080 mol K x 39.10 g/mol

                              = 3.13 g K

Therefore, The 3.13 g of K would be needed to completely react with the remaining.

Example verification:

Suppose we had an additional 0.50 g of \(Cl_2\) in the reaction. Would all of the K be consumed, or would there still be excess K?

Moles of additional \(Cl_2\) = mass of \(Cl_2\) / molar mass of \(Cl_2\)

Moles of additional \(Cl_2\) = 0.50 g / 70.90 g/mol

Moles of additional \(Cl_2\) = 0.0070 mol

The theoretical yield of KCl that can be formed from the additional \(Cl_2\) is:

0.0070 mol \(Cl_2\) x (2 mol KCl / 1 mol \(Cl_2\)) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl

Therefore, the total amount of KCl that can be formed from all of the \(Cl_2\) is:

5.95 g + 1.04 g = 6.99 g

The amount of K that would be needed to completely consume all of the \(Cl_2\).

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Answer: An electron con only exist in a limited number of quantized energy levels.

Explanation:

Answer:

1) d = 2.4 g/cm³

2) m = 25 g

3) v = 126.7 cm³

Explanation:

Given data:

Mass of material = 24 g

Volume of material = 10 cm³

Density of material = ?

Solution:

Formula:

d = m/v

by putting value,

d = 24 g / 10 cm³

d = 2.4 g/cm³

2) Given data:

Density of material = 5 g/cm³

Volume of material = 5 cm³

Mass of material = ?

Solution:

Formula:

d = m/v

5 g/cm³ = m / 5 cm³

m = 5 g/cm³×5 cm³

m = 25 g

3)Given data:

Density of material = 3 g/cm³

Mass of material = 380 g

Volume of material =  ?

Solution:

Formula:

d = m/v

3 g/cm³ = 380 g / v

v = 380 g /3 g/cm³

v = 126.7 cm³

Answer:

4.42 x 10^22 molecules

Explanation:

1 mole of CO2 has 6.022 x 10^23 molecules

=> 0.0734 x 6.022 x 10^23 = 4.420148 x 10^22 or 4.42 x 10^22

Answer:

pH = 5.35

Explanation:

Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.

Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).

      HOAc(aq) ⇄          H⁺(aq)            +                 OAc⁻(aq)

 I      0.10m          1.32 x 10⁻³M ≈ ∅M*     (1.6g/82.03g/mol) / 0.050L = 0.39M

 C       -x                         +x                                0.39M + x ≈ 0.39M**

 E    0.10M - x                  x                                    0.39M

       ≈ 0.10M

Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]

[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M

∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35

_______________________________________________

* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).

** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.

*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.

The pH of the solution measures the acid of the liquid throughout the molar concentration of hydrogen ions in a solution.

\(\bold{CH_3COONa \ mass = 1.60\ g}\\\\\)

Solution Volume (V) \(=50.0\ mL =0.05\ L\\\\\)

\(\bold{CH3COONa \ molarity =0.10\ M }\)

\(\bold{\text{Calculating the CH3COONa moles} =\frac{mass}{molar\ mass}}\)

                                                       \(= \frac{1.60\ g}{82.03\frac{g}{mol}} \\\\=0.0195\ mol\\\\\)

\(\bold{M = \frac{ \text{amount of solute moles} } { \text{solution volume in L} }}\)

     \(= \frac{0.0195\ mol}{0.05\ L}\\\\ =0.39\ M\)

Using Henderson Hasselbach equation:

\(\bold{pH = -\log K_{a} + \log{[salt]}{[acid]}}\\\\\)

     \(=\bold{ -\log (1.75\times 10^{-5}) + \log ( \frac{0.39}{0.10}) }\\\\=\bold{ 4.757 + \log (3.9)}\\\\=\bold{ 4.757 + 0.5910}\\\\=\bold{ 5.348}\\\\=\bold{ 5.35}\\\)

So, the final answer is "5.35".

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The concentration of the HNO₃ solution needed to neutralize the 27.60 mL of 1.00 M NaOH is 2.76 M

The balanced equtaion is given below:

HNO₃ + NaOH —> H₂O + NaNO₃

Now, we shall obtain the concentration of the HNO₃ solution needed for the neutralization reaction. This is shown below:

CaVa / CbVb = nA / nB

(Ca × 10) / (1 × 27.6) = 1

(Ca × 10) / 27.6 = 1

Cross multiply

Ca × 10 = 27.6

Divide both side by 10

Ca = 27.6 / 10

Ca = 2.76 M

Thus, the concentration of the HNO₃ solution needed is 2.76 M

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Answer:

6

Explanation:

Glucose has a chemical formula of: C6H12O6 That means glucose is made of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms.

D. a bond between two atoms that have equal electronegativities

Covalent bonds involve 2 atoms sharing electrons.

Covalent Bonds

There are 3 types of bonds: metallic, ionic, and covalent. Metallic bonds occur between 2 metals that exist in a "sea of electrons." Ionic bonds have high electronegativity differences and occur between a metal and a nonmetal. Finally, as stated above, covalent bonds occur when 2 atoms share their electrons. Covalent bonds usually occur between two nonmetals. However, there are 2 types of covalent bonds: polar and nonpolar.

Nonpolar Bonds

Both polar and nonpolar bonds involve the sharing of electrons; however, polar bonds share electrons unequally. This is caused by an electronegativity difference greater than 0.5. When two atoms have equal electronegativities, they share the electrons equally. This creates a nonpolar bond.  

Hey there!

Molar mass NaCl = 58.44 g/mol

58.44 g ----------------- 22.4 ( at STP )

3.2 g -------------------- Volume ??

Volume = ( 3.2 x 22.4 )  / 58.44

Volume = 71.68 / 58.44

Volume = 1.226 L

Hope this helps!

Answer:

V = 0.56 L

Explanation:

Given data:

Mass of NaCl = 3.2 g

Volume of chlorine gas = ?

Pressure and temperature = standard

Solution:

Chemical equation:

2NaCl    →     2Na + Cl₂

Number of moles of NaCl:

Number of moles = mass/molar mass

Number of moles = 3.2 g/ 58.44 g/mol

Number of moles = 0.05 mol

Now we will compare the moles of NaCl with chlorine;

              NaCl       :          Cl₂

                 2          :           1

                0.05      :       1/2×0.05 = 0.025 mol

Volume of chlorine:

PV = nRT

V = nRT/P

V = 0.025 mol × 0.0821 atm.L/mol.K × 273 K / 1 atm

V = 0.56 L

Answer:

0.1288g of KOH are needed.

Explanation:

1st) It is necessary to write and balance the chemical reaction:

From the balanced reaction we know that 1 mole of KOH neutralizes 1 mole of HCl.

2nd) We have to calculate the number of HCl moles that are contained in 13.4mL of a 0.17M solution.

The molarity of the solution indicates that there are 0.17 moles of HCl in 1000mL of solution, so we can calculate the number of moles in 13.4mL of solution, using a mathematical rule of three:

Now we know that there are 2.3x10^-3 moles of HCl in the solution.

3rd) From the stoichiometry of the reation, 1 mole of KOH neutralizes 1 mole of HCl. So, since the relation between KOH and HCl is 1:1, in this case, 2.3x10-3 moles of KOH are needed to neutralize the HCl.

Using the molar mass of KOH (56g/mol) we can convert the moles into grams:

Finally, 0.1288g of KOH are needed.