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il63 [147K]
3 years ago
8

14. Which graph (below) accurately illustrates the motion of description A:

Chemistry
1 answer:
sertanlavr [38]3 years ago
8 0
I can’t see the picture it’s blank take another willing to help
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The instruction booklet for your pressure cooker indicates that its highest setting is 12.8 psi . You know that standard atmosph
lubasha [3.4K]

Answer:At 237.57^oC food will cook in the pressure cooker set on high.

Explanation:

Standard atmospheric pressure P_1= 14.7 Psi = 1.0001 atm (1 Psi=0.06804 atm)

Standard temperatureT_1 = 273.15 K

Highest pressure offered by pressure cooker: 12.8 Psi = 0.8709 atm

Pressure (P_2) inside the pressure cooker on the highest setting at temperatureT_2 :

P_2 =  1.0001 atm + 0.8709 atm = 1.8710 atm

According to Gay lussac law  :

(pressure)\propto (temperature) (at constant Volume)

\frac{P_1}{T_1}=\frac{P_2}{T_2}

T_2=\frac{P_2\times T_1}{P_1}=\frac{1.8710 atm\times 273.15 K}{1.0001 atm}=510.73 K= 237.57^oC (T(^oC)=T-273.15 K)

At 237.57^oC  food will cook in the pressure cooker set on high.

8 0
3 years ago
Read 2 more answers
What is the balanced chemical equation for caffeine?
Olegator [25]
<span>C8H10N4O2 
Or do you want it in a reaction problem? 


</span>
8 0
3 years ago
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
Where might you find a temperature of 2 Kelvin?
anyanavicka [17]
D. The coldest regions of outer space
5 0
3 years ago
PLEASE HELP!
noname [10]
Your answer should be A and B
7 0
3 years ago
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