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Zarrin [17]
3 years ago
5

Please show all of your work! :)

Chemistry
1 answer:
Paladinen [302]3 years ago
3 0

Answer:

A

Explanation:

To answer this, we need to use Gay-Lussac's law, which states that:

\frac{P_1}{T_1}= \frac{P_2}{T_2} , where P is pressure and T is temperature

The initial pressure we're given is 4.5 atm (so P1 = 4.5) and the temperature is 45.0°C; however, we need to change Celsius to Kelvins, so add 273 to 45.0: 45.0 + 273 = 318 K (so T1 = 318).

The final pressure is what we want to find, but we do know the final temperature is 3.1°C. Converting this to Kelvins, we get: 3.1 + 273 = 276.1 K, which means T2 = 276.1.

Plug these values in:

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

\frac{4.5}{318}= \frac{P_2}{276.1}

Multiply both sides by 276.1:

P_2 ≈ 3.9 atm

The answer is thus A.

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23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
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Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

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3 years ago
What is the role of a consumer in the flow of energy through a food chain?
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1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g
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1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0
3 years ago
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