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2 years ago

In an exothermic reaction, energy is blank the surroundings.

Chemistry

2 answers:

S_A_V [24]2 years ago

5 0

Answer: released to; absorbed from

- In an exothermic reaction, energy is released to the surroundings.

- In an endothermic reaction, energy is absorbed from the surroundings.

Explanation:

An exothermic reaction is a chemical reaction that occurs spontaneously and brings about the release of energy to the surroundings. Hence, the reacting vessel feels hot as the reaction proceeds.

An endothermic reaction, on the other hand, does not occur spontaneously and proceed only when energy is absorbed from the surroundings. Hence, the reacting vessel feels cold as the reaction proceeds.

Art [367]2 years ago

4 0

Answer:in an exothermic reaction, energy is given office into the surroundings while in endothermic reaction energy is absorbed from the surrounding

Explanation:

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The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

2 years ago

When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b

Drupady [299]

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂

3 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

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Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$