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2 years ago

The density of gold is 19.3 g/cm3. what volume would 131.2 g of gold have?

Chemistry

1 answer:

Luda [366]2 years ago

5 0

Volume would be 6.79792764 cm³. Just divide 131.2 by 19.3

ρ=Mass/Volume
Volume= mass/density

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A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperat

stira [4]

Answer:

8,19 L

Explanation:

The combined gas law is a equation that can be used when the initial and final conditions -pressure,volume,amount of moles,temperature-, of a gas change during a process. It can be written:

$\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}$

If the amount of the gas remains constant, then n1=n2 and we have:

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

For the problem we have:

Note: When working with gases is important to use absolute temperature values (°K, °K=°C+273,15):

P1=1 atm, V1=2,5L, T1=25+273,15=298,15°K

P2=0,3 atm, V2=?, T2=20+273,15=293,15°K

$V2=\frac{P1V1T2}{T1P2}=\frac{1atm*2,5L*293,15K}{298,15K*0,3atm}=8,19L$

The new volume of the balloon is 8,19 L.

5 0

2 years ago

Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetic acid in a buffer solution of 0.2M at pH 4.9.

masya89 [10]

Answer:

0.084 M

Explanation:

Using the Henderson-Hasselbalch equation for a buffer ( a buffer is solution contain a weak acid and it conjugate base; the solution resist change in pH)

pH = pKa + log ( base/acid)

4.9 - 4.76 =log ( base / acid)

10^0.14 = ( base / acid)

1.38 = (base / acid)

since there is 0.2 M in the buffer solution

the concentration of acid = $\frac{1}{(1+1.38)}$ × 0.2 = 0.084 M

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2 years ago

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Ivanshal [37]

Answer:

It is pseudoscience because there are at least 118 known elements in nature.

Explanation:

Alchemy is an ancient branch of natural philosophy (not used much anymore).

I also just took the test and it's correct.

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3 years ago

In a solution of salt and water blank is the solvent

Greeley [361]

Water is the solvent

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2 years ago

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san4es73 [151]

Answer: ABC

Explanation:

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3 years ago

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