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ArbitrLikvidat [17]
3 years ago
13

3. What is the gram-formula mass of (NH4)3PO4?

Chemistry
1 answer:
nasty-shy [4]3 years ago
4 0
The gram-formula mass means that the mass of one mole molecule. Then you just need to calculate the elements together. So the gram-formula mass of (NH4)3PO4 is equals to (14+4)*3+31+16*4= 149g/mol. So the answer is (c)
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At 0 degrees Celsius, a gas occupies 22.4L. How hot must the gas be in celcius to reach a volume of 25.0L
NikAS [45]

Answer:

31.7 °C

Explanation:

Charles law states that for volume of a gas is directly proportional to the absolute temperature for a fixed amount of gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at second instance

temperature should be in kelvin scale

T1 - 0 °C + 273 = 273 K

substituting the values in the equation

22.4 L / 273 K = 25.0 L / T2

T2 = 304.7 K

temperature in celcius is - 304.7 K - 273 = 31.7 °C

the gas must be 31.7 °C to reach a volume of 25.0 L

7 0
3 years ago
Is the study of matter and energy complete, or do you think there’s still new information to discover?
Vladimir [108]

Answer:

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Explanation:

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5 0
3 years ago
Can someone help me fill in the rest of the that don't have any words on top of the pictures. Please label the number then the w
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3 0
2 years ago
Where does the equilibrium point occur in a reaction system?
Ymorist [56]
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8 0
3 years ago
What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?
tamaranim1 [39]
<h2>Answer:AlF_{3} </h2>

Explanation:

The chemical equation of the reaction that occurs when AlF_{3} reacts with O_{2} is

4AlF_{3}+3O_{2}→2Al_{2}O_{3}+6F_{2}

4 moles of AlF_{3} requires 3 moles of O_{2}.

1 mole of AlF_{3} requires \frac{3}{4} moles of O_{2}.

Given that we have 9 moles of AlF_{3}.

9 moles of AlF_{3} requires \frac{3}{4}\times 9=6.75 moles of O_{2}.

But we have 12 moles of O_{2}.

So,AlF_{3}  will be consumed first.

So,AlF_{3}  is the limiting reagent.

3 0
3 years ago
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