B. -C- -C- | | | -C-C-C-C- | | | -C- 1 TT -C-

when 45 grams of copper (ii) carbonate are decomposed with heat how many grams of carbon dioxide will be produced

Maksim231197 [3]

Answer:

16.02 g

Explanation:

the balanced equation for the decomposition of CuCO₃ is as follows

CuCO₃ --> CuO + CO₂

molar ratio of CuCO₃ to CO₂ is 1:1

number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol

according to the molar ratio

1 mol of CuCO₃ decomposes to form 1 mol of CO₂

therefore 0.364 mol of CuCO₃ decomposes to form 0.364 mol of CO₂

number of CO₂ moles produced - 0.364 mol

therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g

16.02 g of CO₂ produced

8 0

3 years ago

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

Degger [83]

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

7 0

3 years ago

Read 2 more answers

How many moles of sodium atoms do you have if you have 5.60 ~

Papessa [141]

Answer:

0.93 mol

Explanation:

Given data:

Number of moles of Na atom = ?

Number of atoms = 5.60× 10²³

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

5.60× 10²³ atoms × 1 mol / 6.022 × 10²³ atoms

0.93 mol

3 0

3 years ago

How many of the following molecules have sp2 hybridization on the central atom? hcn so2 ocl2 xecl2.

GrogVix [38]

Only one molecule have sp2 hybridization on central atom and that is SO₂.

XeCl₂ have sp3d hybridization.

OCl₂ have sp3 hybridization.

HCN have sp hybridization.

Hybridization is defined as the concept of mixing two atomic orbitals to give rise to a new type of hybridized orbitals.

Hybridization intermixing usually results in the formation of hybrid orbitals having entirely different energies, shapes, etc.

Learn more about hybridization here:- brainly.com/question/22765530

#SPJ4

7 0

9 months ago

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago