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Georgia [21]
3 years ago
9

The value of δh° for the reaction below is -482 kj. calculate the heat (kj) released to the surroundings when 24.0 g of co (g) r

eacts completely. 2co (g) + o2 (g) → 2co2 (g)
Chemistry
1 answer:
kiruha [24]3 years ago
8 0

The given thermochemical equation is:

2 CO (g) + O_{2}(g) --->2CO_{2}(g) ΔH^{0} = -482 kJ/mol

Mass of CO given = 24.0 g

Moles of CO = 24.0 g CO * \frac{1 mol CO}{28 g CO} = 0.857 mol CO

Finding out the heat released to the surroundings when 24.0 g CO:

0.857 mol CO * \frac{482 kJ}{2 mol CO} =  207 kJ

Therefore, 207 kJ of heat is released to the surroundings when 24.0 g CO reacts completely.

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The law of conversion states that mass is never created or destroyed. Explain how this law requires all chemical equations to be
ikadub [295]

Answer:

See Explanation

Explanation:

The Law of Conservation of Matter as applied to chemical reactions says that matter is neither created nor distroyed, only changed in form. This implies that the mass of substances going into a reaction process must equal the mass of products generated during the reaction process.

Empirically,

∑ mass reactants = ∑ mass products

One can test this idea after balancing a chemical equation by determining the sum of formula weights of reactants and products; then compare. If reaction was properly balanced, the total mass reactants = total mass of products.

Example:

Combustion of Methane =>  CH₄(g) + 2O₂(g)   =>  CO₂(g) + 2H₂O(l)

Equation Weights =>             16amu + 64amu <=> 44amu +  36amu

Mass Reactants = Mass Products  => 80amu <=> 80amu.

__________________

*amu = atomic mass units => sum of atomic weights of elements

3 0
4 years ago
What are the empirical formula and empirical formula mass for C10H30O10?
AysviL [449]

Answer:

Empirical formula: CH₃O

Empirical formula mass = 31 g/mol

Explanation:

Data Given:

Molecular Formula = C₁₀H₃₀O₁₀

Empirical Formula = ?

Empirical Formula mass =

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule

As

C₁₀H₃₀O₁₀ Consist of  10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.

Now

Look at the ratio of these three atoms in the compound

                         C : H : O

                        10 : 30 : 10

Divide the ratio by two to get simplest ratio

                          C      :   H      :    O

                         10/10 : 30/10 : 10/10

                             1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

                     Empirical formula of C₁₀H₃₀O₁₀ =  CH₃O

Now

To find the empirical formula mass in g/mol

Formula mass:

Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.

**Note:

if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol

So,

As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O

Then Its empirical formula mass will be

CH₃O

Atomic Mass of C = 12

Atomic Mass of H = 3

Atomic Mass of O = 16

Total Molar mass of CH₃O

CH₃O = 12 + 3(1) + 16

CH₃O = 12 + 3 + 16

CH₃O = 31 g/mol

4 0
3 years ago
A solution is 40 cetic acid by mass. the density of this solution is 1.049 g/ml. Calculate the mass of pure acetic acid in 220 m
taurus [48]

The mass of pure acetic acid in 220 ml of the given solution at 20°C is 92.311 g

<h3>What is Acetic acid?</h3>

Acetic acid is a type of carboxylic acid and also known as ethanoic acid

Its formula is CH₃COOH.

It is an organic compound and is a colorless liquid

It is mostly used in the production of vinegar

40 % acetic acid by mass means,

40 g of acetic acid is dissolved in 100 g of solution.

The density of solution at 20°C,

\rho = 1.049 g/ml

We know,

\rho = \frac{m}{V}

V = \frac{m}{\rho}

The volume of the solution, V = \frac{100}{1.049} = 95.33 ml

95.33 ml of solution contains 40 g of pure Acetic acid

220 ml of solution contains\frac{40 \times 220}{95.33} = 92.311 g of pure Acetic acid

Thus, the mass of pure Acetic acid in 220 ml of solution at 20°C is 92.311 g

Learn more about acetic acid:

brainly.com/question/24304533

#SPJ4

7 0
2 years ago
How can you figure out the properties of a star?
Firlakuza [10]

Answer:

Measuring distances to stars through stellar parallax. Measuring distances to stars through the moving cluster method. Astronomers measure the temperature of a star by looking at the star's color and its spectrum. The apparent brightness of a star tells how bright it seems to us

Explanation:

3 0
3 years ago
Read 2 more answers
How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2
jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol

For the given chemical reaction:

BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = \frac{1}{1}\times 0.00286mol=0.00286mol of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0
3 years ago
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