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Tasya [4]
3 years ago
12

Which of the following samples will have the greatest volume at STP?

Chemistry
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

Option b. 22 g of He will have the greatest volume at STP

Explanation:

In order to determine the volume, we apply the Ideal Gases Law equation:

P . V = n . R . T

V = n . R . T / P

R, T and P are the same in all the situation we must define n (number of moles).

The one that has the greatest number of moles will have the greatest volume at STP

22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne

22g of He . 1mol / 4 g = 5.5 moles of He

22 g of O₂ . 1mol / 32g = 0.68 moles of O₂

22 g of Cl₂ . 1mol / 70.9 g = 0.31 moles of Cl₂

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4 years ago
What kind of glacier has pieces that breaks off as icebergs
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3 years ago
How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

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6 0
3 years ago
0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
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Answer:

Explanation:

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final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

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No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

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No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

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