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xxTIMURxx [149]
3 years ago
10

Which substance has Delta.Hf defined as 0 kJ/mol?

Chemistry
1 answer:
miskamm [114]3 years ago
6 0

Answer:

the answer is 02 (g)

Explanation:

hope this helps :)

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Kay [80]
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8 0
3 years ago
A sample of methane (CH4) has a volume of 25 mL at a pressure of 0.80 atm. What is the volume of the gas at each of the followin
Xelga [282]

Answer:

a. 50ml b.10ml c. 6.097ml d. 190.1 ml

Explanation:

According to Boyle's law

Volume is inversely proportional to pressure at constant temerature

Mathematically

P1V1=P2V2

P1=Initial pressure=0.8atm

V1=Initial volume=25ml

making V2 the subject

at 0.4atm P2=0.4 atm,

V2=25×0.8/0.4

=50ml

at 2 atm V2=25×0.8/2

=10 ml

1mmHg=0.00131579

2500mmHg=3.28 atm

At 3.28 atm,V2=25×0.8/3.28

=6.097 ml

at 80.0 torr

1 torr=0.00131579

80 torr=0.1052 atm

at 0.1048 atm V2=25×0.8/0.1048

=190.1 ml

4 0
3 years ago
Which of the following is true about two neutral atoms of the element gold? (1 point) The nucleus is missing in both. Each has a
klio [65]

i have the same question idk the anwser tho

5 0
3 years ago
According to each substances heat of fusion, which of the items below requires more heat to be added per gram of substance to go
alexira [117]

<u>Answer: </u>The correct answer is Silver.

<u>Explanation:</u>

Specific heat of fusion is defined as the amount of heat which is required to raise the temperature of 1 gram of a substance to 1°C. It is generally expressed in kJ/mol

We are required to find the substance which require more heat. For that we need to know the specific heat of all the substances.

The substance which have the highest specific heat, will require more heat.

The specific heat of the given substances are:

Silver = 11.3 kJ/mol

Sulfur = 1.7175 kJ/mol

Water = 5.98 kJ/mol

Lead = 4.799 kJ/mol

The specific heat of silver is the highest and hence, will require more heat.

Hence, the correct answer is silver.

7 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
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