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4 years ago

A sample of hydrogen gas is found to be 8.50 L at a temperature of 38.0 C and a pressure of 725mm Hg. Find the volume at STP.

1 answer:

6 0

Temp must be Kelvin
38 C = 311.15 K
Volume at STP = 8.50 liters * (273.15 / 311.15) * (725 / 760) =
 7.1182746306 Liters

The formula to use is:

Volume at STP = Present Volume * (273.15 / Present Temp °K) * (Present Pressure (Torr) / 760)

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Determine how many liters 8.65 g of carbon dioxide gas would occupy at the following conditions.

VashaNatasha [74]

1) 6.17 liters

2)2.66 liters

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3 years ago

What is the molarity of a nitric acid solution if 43.13 mL 0.1000 M KOH solution is needed to neutralize 30.00 mL of the acid so

Leno4ka [110]

Answer:

0.144M

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

HNO3 + KOH —> KNO3 + H20

From the equation,

nA = 1

nB = 1

From the question given, we obtained the following:

Ma =?

Va = 30.00mL

Mb = 0.1000M

Vb = 43.13 mL

MaVa / MbVb = nA/nB

Ma x 30 / 0.1 x 43.13 = 1

Cross multiply to express in linear form

Ma x 30 = 0.1 x 43.13

Divide both side by 30

Ma = (0.1 x 43.13) /30 = 0.144M

The molarity of the nitric acid is 0.144M

8 0

3 years ago

How many moles of water contain 1.3 x 10^24 molecules?

Vika [28.1K]

C is the correct answer

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4 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

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3 years ago

Describe examples of body language that show that you are actively listening

stellarik [79]

Answer:

Explanation:

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3 years ago