An atom of nitrogen has several protons in its nucleus. How many electrons does it have

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 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.

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

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

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

54.0g Al reacts with 64.0g O2 to form Al2O3 according to the equation.

pentagon [3]

Answer:

136 g Al₂O₃

Explanation:

Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).

Molar Mass (O₂): 32 g/mol

Molar Mass (Al₂O₃): 102 g/mol

4 Al + 3 O₂ -----> 2 Al₂O₃

64.0 g O₂ 1 mole 2 moles Al₂O₃ 102 g
----------------- x -------------- x ------------------------ x ------------- = 136 g Al₂O₃
32 g 3 moles O₂ 1 mole

6 0

2 years ago

Read 2 more answers

When a 10 ml graduated cylinder is filled to the 10 ml mark, the mass of the water was measured to be 9.955 g. if the density of

Lady_Fox [76]

Given, the density of water is 0.9975 g/ml. Density of water is mass of water per unit volume. Mass of 1 ml of water supposed to be 0.9975 g from density of water. So, mass of 10 ml of water is (0.9975 X 10) g= 9.975 g. From graduated cylinder, mass of 10 ml water is measured to be 9.955 g. So, error for mass of 10 ml water= (9.975-9.955)=0.02 g. Percentage of error for 10 ml water is $\frac{0.02}{10} X 100$ = 0.2. Error in the mass for the 10 ml of water is 0.2 %.

5 0

3 years ago

I don't understand

inysia [295]

Answer:

A 9 moles of reactants changes into 11 moles of product

Explanation:

8 0

3 years ago

Gekata [30.6K]

Both Chemical and Nucleus Reactions form new substances.
Mark me as brainliest if I helped:)

4 0

3 years ago