What is (-1.7588 × 10^11 C/Kg)(1.309 × 10^-15 Kg)

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=11.3L\\T_1=32^oC=(32+273)K=305K\\V_2=?\\T_2=-5^oC=(-5+273)K=268K$

Now put all the given values in above equation, we get:

$\frac{11.3L}{305K}=\frac{V_2}{268K}\\\\V_2=9.93L$

Therefore, the volume of gas sample will be, 9.93 L

6 0

3 years ago

Mr. Holmes is busy making a paper mache volcano that he is sure is going to win the national science fair contest. He only has r

AnnyKZ [126]

Answer:

Bottle 2 shows the vinegar has a molarity of 2M acetic acid

Explanation:

2M of acetic acid has higher concentration of acetic acid as compared to other vinegar. In 2M of acetic acid is equals to 120 gram of acetic acid in 500 ml of water which is quiet higher than other concentration of acetic acid in vinegar which is 4%. So Mr. Holmes has to use 2M of acetic acid which will produced high intense eruption in the volcano and win the national science fair contest.

7 0

3 years ago

The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction

shepuryov [24]

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

$NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)$

(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of $H_{2}S$ will increase.

(b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of $H_{2}S$ will decrease with decrease in volume.

When we increase the mount of $NH_{4}HS$ then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of $H_{2}S$ will increase then.

3 0

3 years ago