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Luden [163]
3 years ago
14

A(n) 816 ml sample of hydrogen was collected over water at 21.0◦c on a day when the barometric pressure was 755 torr. what volum

e would the dry hydrogen occupy under standard conditions? the vapor pressure of water at 21.0◦c is 19.0 torr. answer in units of ml.
Chemistry
1 answer:
hammer [34]3 years ago
6 0
The correct answer is A

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vladimir2022 [97]
To fill the other element's shell.
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3 years ago
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2) A balloon was inflated to a volume of 5.0 liters at a temperature of
TEA [102]

Answer:

6.12 L

Explanation:

Given that,

Initial volume, V₁ = 5 L

Initial temperature, T₁ = 7.0°C = 343 K

Final temperature, T₂ = 147°C = 420 K

We need to find its new volume. The relation between volume and temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{5\times 420}{343}\\\\V_2=6.12\ L

So, the new volume is 6.12 L.

8 0
2 years ago
Which statement correctly pairs the climate factor with its effect on temperature? *
allochka39001 [22]

Answer:

A region on top of a mountain is cooler than at the base.

Explanation:

Pressure and temperature have direct relationship with each other. With the decrease in pressure, the temperature decreases and vice versa. When the air rises in the atmosphere, the pressure starts to fall. The low pressure at the peak of the mountains tends to cause the fall in temperature. It is because of this reason that it is cooler at the top of the mountain while the temperature is less cool in the foothills.

5 0
2 years ago
What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C
Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = \frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\

Now use the molar ratio to find how many moles of NaOH would be required to neutralize 5.5537*10^{-3} mol\\ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

5.5537*10^{-3} mol\\ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L

5 0
3 years ago
5. Water, wind, ice, and gravity are important agents of ___ which is a destructive force​
Julli [10]
I agree with the statement that the other person has made
6 0
2 years ago
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