where the solubility rules completely adequate

The equation must have the following form to be balanced: 4CO2 + 2H2O = 2C2H2 + 5O2

S8: O2 = 1/8; S8 utilized = 240/8 = 30SO2 generated is equal to the number of sulphur molecules used, which is 240. Stoichiometry is the name given to the study of chemical processes in mathematics. Numerous calculations can be done, such as stoichiometry, which is most usually performed with moles but can also be done with masses and even percentages. Stoichiometric ratio A stoichiometric ratio is important when considering the interactions between specific elements or molecules. This exact ratio of reactant to product coefficients is necessary for a reaction to occur properly. Let's discuss some problems you can run across when you learn about stoichiometry. Chemical Equations in Balance Equations needing to be balanced is a fairly common stoichiometric issue type. This is an essential chemistry skill since a reaction can only occur if the ratio of reactants to products is correct.possess. Additionally, it provides an essential framework for organic chemistry. Balance the ensuing reply: _ CO2 + _ H2O C2H2 + _ O2 To be balanced, equations must have an equal number of each element on both sides of the reaction. Before balancing the oxygen, you can start by balancing the carbons and hydrogens. The equation must have the following form to be balanced: 4CO2 + 2H2O = 2C2H2 + 5O2

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Answer:

\(\boxed {\boxed {\sf \rho= 0.1 \ g/cm^3 }}\)

Explanation:

Density is found by dividing the mass by the volume.

\(\rho=\frac{m}{v}\)

The mass is 50 grams, but we have to find the volume.

Assuming this is a solid rectangular object, we can use this formula for volume:

\(v=l*w*h\)

The length is 25 centimeters, the width is 10 centimeters, and the height is 2 centimeters.

\(v= 25 \ cm * 10 \ cm * 2 \ cm\)

\(v= 250 \ cm^2 * 2 \ cm\)

\(v= 500 \ cm^3\)

Now we know the mass and volume:

\(m= 50 \ g\\v= 500 \ cm^3\)

Substitute the values into the formula.

\(\rho = \frac{50 \ g}{500 \ cm^3}\)

\(\rho= 0.1 \ g/cm^3\)

The density of the object is 0.1 grams per cubic centimeter.

Answer:

they both move at a different force... believe I'm right

Explanation:

Answer:

The moon

Explanation:

Force? I think

Answer:

i think your answer is d  

Explanation:

sorry if it is wrong

Answer:

Ingredients OR tenderising

Explanation:

Answer:

14

Explanation:

If the sum of three consecutive even numbers is 48, the smallest of these numbers is 14.

First, we need to calculate the heat absorbed by the water. Next, we need to use the principle of conservation of energy to calculate the heat released by the pan.

We can use the formula:

Q = m * c * ΔT

We have:

m = 2.0 kg (mass of water)

c = 4.18 kJ/kg·°C (specific heat capacity of water)

ΔT = (63.2°C - 25.0°C) = 38.2°C

Q = (2.0 kg) * (4.18 kJ/kg·°C) * (38.2°C) = 319.22 kJ

Next, we need to use the principle of conservation of energy to calculate the heat released by the pan. We can assume that all the heat released by the pan is absorbed by the water and the calorimeter. We can use the formula:

Q = m * c * ΔT

We have:

m = 3.0 kg (mass of the pan)

c = ? (specific heat capacity of the pan)

ΔT = (300.0°C - 63.2°C) = 236.8°C

Q = 319.22 kJ

Substituting the values, we get:

319.22 kJ = (3.0 kg) * c * (236.8°C)

Solving for c, we get:

c = 319.22 kJ / (3.0 kg * 236.8°C) = 0.449 kJ/kg·°C

The metal that the pan is made of is probably cast iron in this case.

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Answer:

qwertuyilolassddfhhjkllzxcvbnnnmetyio

Answer:

c

Explanation:

metals are good conductors, while non metals are not good conductors

Answer:

B

Explanation:

it is the energy stored in the chemical bonds of a substance

Answer:

B. The energy of position or composition

Explanation:

Potential energy is the energy possessed by an object or a body due to it's position in relation with other objects or bodies. It is a stored energy.

Answer:

1/4

Explanation:

0.25 is equal to 1/4

The effect of errors on the calculated concentration of CI is significant.

A 10.00 mL diluted chloride sample was titrated with 0.02749 M AgNO3, and 16.51 mL AgNO, was required to reach the endpoint. The effect of errors on the calculated concentration of CI can be explained as follows:a. The student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M. If the student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M, then the experimentally calculated moles of Ag would be too high. Consequently, the calculated [CI] in the unknown would come out too low. b.

The student was past the endpoint of the titration when the final buret reading was taken. If the student was past the endpoint of the titration when the final buret reading was taken, then the experimentally determined moles of Ag would be too low. This would cause the calculated C1 concentration to come out too high. Consequently, the calculated moles of CI would come out too high. Therefore, the effect of errors on the calculated concentration of CI is significant.

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Some of the things that might be included in a chemical reaction are: reactants, products, catalysts, energy, and reaction conditions.

A chemical reaction involves the breaking and forming of chemical bonds to create new substances. Some of the things that might be included in a chemical reaction are:

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Answer: northern hemisphere

Explanation: I looked it up. Plus I took a test with this question and when the teacher went over the answer i got it right.

Answer:

Explanation:

no. of moles of butane = 5.87g / 58.124 = 0.1009mol

since O2 is excess, limiting reagent is butane

from the equation, 2 moles of butane will reacted to produce water = 10 moles

therefore, 0.1009 moles of butane produced water = (0.1009 x 5) = 0.5050 moles

Weight of H2O formed = 0.5050 x 18.015 = 9.10 grams

Freezing point of a solution containing 1.25 g1.25 g of benzene (C6H6)(C6H6) in 100 g100 g of chloroform (CHCl3)(CHCl3) is -64.249°C.

To register a frozen edge for a plan, you must use the recipe for frozen edges for discouragement.

ΔTf = Kf * molality

where ΔTf is the freezing edge for despair and Kf is the freezing edge for despair. As can be expected for soluble substances, molality is the molal mixture of solutes.

First we have to deal with the molality of the plan.

molality = moles of solute/mass of solute (kg)

The mass of chloroform is 100 g, which is equivalent to 0.1 kg.

molarity = 0.016 mol/0.1kg = 0.16 mol/kg

The reliable freezing point of chloroform for rejection of

is 4.68 K*kg/mol.

ΔTf = 4.68K*kg/mol * 0.16mol/kg = 0.749 K

The freezing limit of pure chloroform is 63.5°C. So the game plan frozen edge is frozen edge = - 63.5°C - 0.749°C = - 64.249°C

So the freezing point of the plan is -64.249°C.

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The mass of 7,00 × 10²² NaOH molecules is 4,64 gram. 7,00 × 10²² NaOH molecules equal 0.116 moles NaOH.

To find out the mass of 7,00 × 10²² NaOH molecules you can use the following steps

Step 1: The first step is to calculate the number of moles of the compound.

mol = number of particles ÷ Avogadro's number

      = 7,00 × 10²² ÷ 6,022 × 10²³

      = 0,116 mol

Step 2: The next step is to calculate the mass of the molecules.

mass = mol × relative molecular mass

         = 0,116 mol × 40 gram/mol

          = 4,64 gram

So the mass of 7,00 × 10²² NaOH molecules is 4,64 gram.

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Answer:

The rate reduces by a factor cf /2.

Answer:

5.5 × 10²² formula units Al₂O₃

General Formulas and Concepts:

Chemistry - Stoichiometry

Explanation:

Step 1: Define

9.3 g Al₂O₃

Step 2: Define conversions

Avogadro's Number

Molar Mass of Al - 26.98 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al₂O₃ - 2(26.98) + 3(16.00) = 101.96 g/mol

Step 3: Convert

\(9.3 \ g \ Al_2O_3(\frac{1 \ mol \ Al_2O_3}{101.96 \ g \ Al_2O_3} )(\frac{6.022 \cdot 10^{23} \ formula \ units \ Al_2O_3}{1 \ mol \ Al_2O_3} )\) = 5.4928 × 10²² formula units Al₂O₃

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

5.4928 × 10²² formula units Al₂O₃ ≈ 5.5 × 10²² formula units Al₂O₃

The value of b in the equation \(\( f(t) = 85 \cdot b^t \)\) represents the decay factor of the radioactive substance. To determine the value of \( b \), we can use the information that the half-life of the substance is 21 years.

The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, the half-life is 21 years, which means that after 21 years, the amount of the substance remaining will be half of the initial amount.

We can use this information to set up an equation:

\(\(\frac{1}{2} = b^{21}\)\)

To solve for b, we need to take the 21st root of both sides of the equation:

\(\(b = \left(\frac{1}{2}\right)^{\frac{1}{21}}\)\)

Using a calculator, we can evaluate this expression:

\(\(b \approx 0.965\)\)

Therefore, the value of b in the equation \(\( f(t) = 85 \cdot b^t \)\) is approximately 0.965.

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The number of moles of PbSO₄(s) precipitated from the resulting solution would be 0.0141 moles.

A precipitation reaction will take place in which the Na₂SO₄(aq) and Pb(NO₃)₂(aq) will react and form PbSO₄(s) solid and NaNO₃(aq).

This is the balanced chemical reaction that takes place:Na₂SO₄(aq) + Pb(NO₃)₂(aq) → PbSO₄(s) + 2NaNO₃(aq)

We first need to determine the number of moles of Na₂SO₄(aq) that is available:0.0200 L × 0.00187 mol/L = 3.74 × 10⁻⁵ mol Na₂SO₄(aq)

Since the reaction has a 1:1 molar ratio between Na₂SO₄(aq) and PbSO₄(s), the number of moles of PbSO₄(s) that will form will be the same.

Therefore, 3.74 × 10⁻⁵ mol PbSO₄(s) will form.In order to calculate the mass of PbSO₄(s) that will precipitate out, we can use the formula:m = n × MM

where m = mass in grams, n = number of moles, and MM = molar mass of PbSO₄The molar mass of PbSO₄ is:1 Pb + 1 S + 4 O = 207.2 g/molSo, mass of PbSO₄(s) = 0.00775 g

We can use the solubility product constant (Ksp) to determine if all of the PbSO₄(s) will precipitate out.Ksp = [Pb²⁺][SO₄²⁻] = 2.5 × 10⁻⁸[Pb²⁺] = [SO₄²⁻] = xMoles of Pb²⁺ and SO₄²⁻ = 0.0141 mol

The molarity of PbSO₄(s) is thus:0.0141 mol ÷ 0.051 L = 0.276 M

This is greater than the Ksp of 2.5 × 10⁻⁸, so not all of the PbSO₄(s) will precipitate out.

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Answer: The amount of CO2 found in the atmosphere varies over the course of a year. Much of this variation happens because of the role of plants in the carbon cycle

Answer:

The bond order is 1.091

Explanation:

there really is no explanation.

Answer:

Explanation:

Isotope notation is of the form

A

Z

X

where  

Z

is the atomic number (number of protons),  

A

is the mass number (sum of protons and neutrons), and   X   is the element.

In Carbon- 14 ,   14  is the mass number (sum of protons and neutrons). This is our  A .

Carbon has the atomic symbol   C , and it has  6 protons. This is the atomic number, or  A . Thus, the isotope notation for Carbon- 14  is  14  6  C

Hope this helps!

Explanation:

Answer:

Phycical science is the study of the body.

Explanation:

Answer:

- 143.6 g of N₂ ; 3.08×10²⁴ molecules of N₂

- 451.4 g of CO₂ ; 6.17×10²⁴ molecules of CO₂

- 153.9 g of H₂O ; 5.14×10²⁴ molecules of H₂O

- 27.3 g of O₂ ; 5.14×10²³ molecules of O₂

Explanation:

This is a reaction of decomposition:

4C₃H₅N₃O₉(s)  →  6N₂(g)  +  12CO₂(g)  +  10H₂O(g)  +  O₂(g)

4 moles of solid nitroglycerine decompose to 6 moles of nitrogen, 12 moles of carbon dioxide, 10 moles of water vapor and 1 mol of oxygen

We convert mass to moles → 777 g. 1mol/ 227g = 3.42 moles

4 moles of C₃H₅N₃O₉ can decompose to:

6 moles of N₂ ____  12 moles of CO₂ ___ 10 moles of H₂O ___ 1 mol of O₂

Then, 3.42 moles of C₃H₅N₃O₉ may decompose to:

(3.42 . 6) / 4  =  5.13 moles of N₂

(3.42 . 12) / 4 = 10.26 moles of CO₂

(3.42 . 10) / 4 = 8.55 moles of water vapor

(3.42 . 1) / 4 = 0.855 moles of oxygen

We convert the moles to mass:

5.13 mol . 28 g/mol = 143.6 g of N₂

10.26 mol . 44 g/mol = 451.4 g of CO₂

8.55 mol . 18 g/mol = 153.9 g of H₂O

0.855 mol . 28 g/mol = 27.3 g of O₂

We count the atoms:

5.13 mol . 6.02×10²³ molecules /mol = 3.08×10²⁴ molecules of N₂

10.26 mol . 6.02×10²³ molecules /mol = 6.17×10²⁴ molecules of CO₂

8.55 mol . 6.02×10²³ molecules /mol = 5.14×10²⁴ molecules of H₂O

0.855 mol . 6.02×10²³ molecules /mol = 5.14×10²³ molecules of O₂

Answer: S waves travel through solids or very dense, glassy liquids. P waves travel through solids liquids and gasses

Explanation:

I took the test, hope this helps!!

Answer:

384.9 mL.

Explanation:

What is given?

Volume 1 (V1) = 823.7 mL.

Pressure 1 (P1) = 351.8 torr.

Pressure 2 (P2) = 752.9 torr.

What do we need? Volume 2 (V2).

Step-by-step solution:

This is a Boyle's Law problem. Boyle's law states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases.

The formula of the law is:

Where P is pressure and V is volume. We want to find volume 2, 'V2'. So let's solve for 'V2' and replace the given data that we have in the new formula:

The new volumewill be 3384.9 mL You can note as the pressure is increasing, the volume is decreasing.

Answer:

it will be b because it is the atom oxygen

Answer:D

Explanation:

I just did it.Brainlist please