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2 years ago

Consider the reaction cacn2 3 h2o → caco3 2 nh3 . how much nh3 is produced if 187 g of caco3 are produced?

1 answer:

7 0

B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.

mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)

mass nh3 = 63.58 g nh3 is produced</span>

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A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

5 0

2 years ago

Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$C(s)+2H_2(g)\rightarrow CH_4(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ$

Hence, the enthalpy of the reaction for the production of $CH_4$ is coming out to be -74.9 kJ

3 0

2 years ago

A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Tanzania [10]

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0

2 years ago

Which of the following statements correctly relates mutations and survival rates of plants?

Tasya [4]

Answer is: <span>Mutations sometimes improve the chances of survival for a plant.
</span>Mutations are very important because they change <span>variability in populations and in that way enable evolutionary change.
</span>There are three types of mutations:
1) good or advantageous mutations - <span> improve the chances of survival for a plant.
2) </span>bad or deleterious - decrease the chances of survival for a plant.
3) neutral - not affect he chances of survival for a plant.

3 0

2 years ago

The shape of an atomic orbital is associated with

Diano4ka-milaya [45]

The properties of the atomic orbital are actually dependent on the quantum numbers.

size of atomic orbital: governed by the principal quantum number (n)

shape of atomic orbital: governed by the angular momentum quantum number (l)

orientation in space: governed by the magnetic quantum number (ml)

Since we are asked about the shape, hence the correct answer is:

angular momentum quantum number (l)

8 0

2 years ago