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4 years ago

Please help this is due tomorrow!

Chemistry

1 answer:

lbvjy [14]4 years ago

7 0

I had this same question on one of are worksheets is D I can’t really explain it but the answer is D

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The rate of disappearance of HBr in the gas phase reaction 2 HBr(g) → H2(g) + Br2(g) is 0.140 M s-1 at 150°C. The rate of appear

djverab [1.8K]

Answer: The rate of appearance of $Br_2$ is $0.0700Ms^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of HBr = $-\frac{1d[HBr]}{2dt}Rate in terms of appearance of [tex]H_2$ = $\frac{1d[H_2]}{dt}$

Rate in terms of appearance of $Br_2$ = $\frac{1d[Br_2]}{dt}$

$-\frac{1d[HBr]}{2dt}=\frac{d[H_2]}{dt}=\frac{d[Br_2]}{dt}$

Given :

$-\frac{1d[HBr]}{dt}=0.140Ms^{-1}$

The rate of appearance of $Br_2$ ;

$\frac{1d[Br_2]}{dt}=-\frac{1d[HBr]}{2dt}=\frac{1}{2}\times 0.140=0.0700Ms^{-1}$

Thus rate of appearance of $Br_2$ is $0.0700Ms^{-1}$

6 0

3 years ago

Question 2 how many valence electrons are in a neutral atom of silicon

marshall27 [118]

Si -Atomic number -14 -1s2 2s2 2p6 3s2 3p2 - 4 valence electrons

7 0

3 years ago

2. What is the approximate mass of an atom that

ratelena [41]

Answer:

b. 45 u

Explanation:

the mass of an atom is equal to the protons plus neutrons:

26+19=45

hope this helps :)

3 0

3 years ago

Read 2 more answers

When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample

andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

$Q=m\times c\times \Delta T$

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

$\Delta T$ = change in temperature = $1.643^oC$

Now put all the given value in the above formula, we get:

$26380J/g=0.235g\times c\times 1.643^oC$

$c=68323.38J/^oC$

Thus, the specific heat of calorimeter is $68323.38J/^oC$

Now we have to calculate the heat of combustion of caffeine.

Formula used :

$Q=c\times \Delta T$

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = $68323.38J/^oC$

$\Delta T$ = change in temperature = $1.584^oC$

Now put all the given value in the above formula, we get:

$Q=68323.38J/^oC\times 1.584^oC$

$Q=108224.23J=108.2kJ$

Now we have to calculate the moles of caffeine.

$\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}$

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

$\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol$

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

$\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole$

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

4 0

3 years ago

Cual es el carbono de valencia libre

Ilia_Sergeevich [38]

Answer:

Explanation:

1) Se escoge la cadena con el mayor número de átomos de carbono. 2) Cuando hay dos cadenas posibles con el mismo número de átomos de carbono, se elige la que contenga el mayor número de triples ligaduras posibles. ... El átomo de carbono con la triple valencia libre se numera como 1.

8 0

3 years ago

Read 2 more answers