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Vaselesa [24]
4 years ago
8

Consider the reaction 2 H 2 + O 2 ⟶ 2 H 2 O Δ H rxn = − 484 kJ Which answer best describes the transfer of heat that occurs when

1.60 mol H 2 reacts? 387 kJ absorbed 484 kJ released 387 kJ released 774 kJ absorbed 484 kJ absorbed 774 kJ released
Chemistry
1 answer:
9966 [12]4 years ago
6 0

Answer: 387 kJ energy is released

Explanation:

Endothermic reactions: These are the reactions in which energy is required in the reaction. The energy of the products are more than the energy of the reactants.\Delta H is positive.

Exothermic reactions: These are the reactions in which energy is released in the reaction. The energy of the products is less than the energy of the reactants.\Delta H is negative.

The given balanced equation is ;

2H_2+O_2\rightarrow 2H_2O  \Delta H=-484kJ

According to stoichioetry :

2 moles of hydrogen reacts to give = 484 kJ of heat

Thus 1.60 moles of hydrogen reacts to give =\frac{484}{2}\times 1.60=387kJ

Thus 387 kJ energy is released

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A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
What mass (g) of barium iodide is contained in 188 ml of a barium iodide solution that has an iodide ion concentration of 0.532m
Katarina [22]

Answer:

What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?

A) 19.6

B) 39.1

C) 19,600

D) 39,100

E) 276

The correct answer to the question is

B) 39.1  grams

Explanation:

To solve the question

The molarity ratio is given by

188 ml of 0.532 M solution of iodide.

Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles

To find the mass, we note that the Number of moles = \frac{Mass}{Molar Mass} from which we have

Mass = Number of moles × molar mass

Where the molar mass of Barium Iodide = 391.136 g/mol

= 0.100016 moles ×391.136 g/mol = 39.12 g

8 0
3 years ago
A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,
Ad libitum [116K]
The answer is 3.68 L
8 0
3 years ago
Read 2 more answers
If you could help me that would be awesome
elixir [45]
There are 3 significant figures. Significant numbers are the numbers that build up your total number. 1-9 always count, 0 only counts if it’s after another number. For example: 0,901 has 3 significant numbers as does 0,910. 9,10 also has 3. 0,09 has just 1.
8 0
3 years ago
Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.
Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O.

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O

Number of atoms present on reactant side are as follows.

  • C = 1
  • H = 4
  • O = 2

Number of atoms present on product side are as follows.

  • C = 1
  • H = 2
  • O = 3

To balance this equation, multiply O_{2} by 2 on reactant side. Also, multiply H_{2}O by 2 on product side. Hence, the equation can be rewritten as follows.

CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O

Now, the number of atoms present on reactant side are as follows.

  • C = 1
  • H = 4
  • O = 4

Number of atoms present on product side are as follows.

  • C = 1
  • H = 4
  • O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O.

4 0
3 years ago
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