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Sonja [21]
3 years ago
12

Measurements show that the enthalpy of a mixture of gaseous reactants increases by 215. kJ during a certain chemical reaction, w

hich is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -155. kJ of work is done on the mixture during the reaction.
Calculate the change in energy of the gas mixture during the reaction. Round your answer to 2 significant digits.


Is the reaction exothermic or endothermic?
Chemistry
1 answer:
Maslowich3 years ago
6 0

Answer:

1. The change in energy is 60KJ or 6.0 × 10^1 KJ

2. Endothermic reaction

Explanation:

The Enthalpy (ΔU) for the mixture is given as 215kJ and the workdone (W) on the mixture is - 155KJ. Hence, the change in the energy (ΔH) ofthe mixture is computed using the equation below:

ΔH = ΔU + w

Where,

ΔH= Change in energy

ΔU= Enthalpy change

W= workdone

Therefore ΔH is:

ΔH= 215 kJ + (-155Kj)

ΔH= 60 KJ

Therefore, the change in energy is 60KJ or 6.0 × 10^1 KJ

The value is positive so it is an endothermic reaction.

An endothermic reaction happens when the energy used to break the bonds in the reactants is higher than the energy given out when bonds are formed in the products. This means that the entire reaction takes in energy, hence there is a temperature decrease in the surroundings. Endothermic reactions cannot happen spontaneously. Work is usually done in order to get these reactions to occur. When endothermic reactions absorb energy, a temperature drop is measured by the reaction.

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The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
Would precipitation occur when 500 mL of a 0.02M solution of AgNO3 is mixed with 500 mL of a 0.001M solution of NaCl? Show your
Oksanka [162]
We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate
7 0
3 years ago
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