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3 years ago

When heated, calcium carbonate decomposes according to the equation given below:

Chemistry

1 answer:

Margaret [11]3 years ago

5 0

Answer:

Percent yield = 90.9%

Explanation:

Given data:

Mass of CaCO₃ = 50.0 g

Mass of CO₂ produced = 20.0 g

Percent yield = ?

Solution:

Chemical equation:

CaCO₃ → CaO + CO₂

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 50.0 g/ 100.1 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of CO₂ with CaCO₃.

CaCO₃ : CO₂

1 : 1

0.5 : 0.5

Mass of CO₂: Theoretical yield

Mass = number of moles × molar mass

Mass = 0.5 mol × 44 g/mol

Mass = 22 g

Percent yield:

Percent yield = ( actual yield / theoretical yield ) × 100

Percent yield = (20.0 g/ 22.0 g) × 100

Percent yield = 0.909 × 100

Percent yield = 90.9%

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Why is the structure of a molecule important to its function

Scilla [17]

Answer:

It determines how biological molecules recognize and respond to one another with specificity.

Explanation:

A molecule has a characteristic size and shape. The precise shape of a molecule is usually very important to its function in the living cell. Molecular shape is crucial in biology because it determines how biological molecules recognize and respond to one another with specificity.

3 0

4 years ago

In a reaction vessel, 17.6 g of solid chromium(III) oxide, Cr2O3, was allowed to react with excess carbon tetrachloride in the f

notsponge [240]

Answer:

72.53% is the yield of CrCl3

Explanation:

Given

Reaction:

Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)

CCl4 is in excess and 17.6g Cr2O3 present

The reaction yields 26.6g of CrCl3

To Find:

% yields of the reaction

Also given

Molar mass of CrCl3 = 158.35g/mol

Molar mass of Cr2O3 = 152.00 g/mol

By the stoichiometry of the reaction

1 mole of Cr2O3 gives 2 moles of CrCl3

1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3

= 1 52 g of Cr2O3 gives 316.70 g of CrCl3

17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3

= 36.67 g CrCl3

but actual yield is only 26.6g

so % yield is (26.6 ÷÷ 36.67) × 100

= 72.53% is the yield of CrCl3

8 0

4 years ago

The decomposition of 57.0 g of Fe2O3 results in Consider the following reaction. 2Fe2O3 ---> 4Fe + 3O2 deltaH degree rxn = +

lozanna [386]

Answer:

The correct answer is option E.

Explanation:

$2Fe_2O_3\rightarrow 4Fe + 3O_2 \Delta H^o_{rxn} = 824.2 kJ$

Mass of ferric oxide decomposed = 57.0 g

Moles of ferric oxide decomposed = $\frac{57.0 g}{160 g/mol}=0.35625 mol$

According to reaction, 2 moles ferric oxide on combustion gives 824.2 kJ of heat.

Then heat given by on decomposition of 0.35625 mol of ferric oxide will be:

$\frac{824.2 kJ}{2}\times 0.35625 =146.81 kJ\approx 147 kJ$

So, on decomposition of 57.0 grams of ferric oxide 147 kilo Joules of heat was released.

3 0

3 years ago

At a different temperature (this means that Keq will be different than part a)), 6.0 mol of IF5 and 8.0 mol of I4F2 are placed i

antoniya [11.8K]

Answer:

Keq for the new temperature is 26.8

Explanation:

Let's propose the equilibrium:

2IF₅ + I₄F₂ ⇄ 3I₂ + 6F₂

Now we propose the situations:

2IF₅ + I₄F₂ ⇄ 3I₂ + 6F₂

Initial 6 mol 8 mol - -

Initially we added 6 mol and 8 mol of our reactants

React. x x/2 3/2x 3x

By stoichiometry x amount has reacted, so a half of x react to the I₄F₂ and we finally produced 3/2x and 3x in the product side

Eq. (6 - x) (8 - x/2) 3/2x 3x

Notice we have the concentration left for the I₄F₂, so we can find the x value, the amount that has reacted:

8 - x/2 = 6

x = 4, so the concentrations in the equilibrium are:

2 moles of IF₅, 6 moles I₄F₂, 6 moles of I₂ and 12 moles of F₂

As we need molar concentration to determine Keq, we must divide the moles by the volume of the container:

2/10 = [IF₅] → 0.2 M

6/10 = [I₄F₂] → 0.6 M

6/10 = [I₂] → 0.6 M

12/10 = [F₂] → 1.2 M

Let's make, expression for Keq:

Keq = ([I₂]³ . [F₂]⁶) / [IF₅]² . [I₄F₂]

Keq = 0.6³ . 1.2⁶ / 0.2² . 0.6 → 26.8

4 0

3 years ago

Which statement is an example of a scientific hypothesis?

Ivanshal [37]

Answer:

Explanation:

5 0

3 years ago