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kicyunya [14]
3 years ago
14

Calculate the mass of butane needed to produce 60.6 g of carbon dioxide.

Chemistry
1 answer:
12345 [234]3 years ago
3 0

Answer:

You take the mass of carbon dioxide, 56.8g, divide by its molar mass, 44.01g/mol, to produce the moles of carbon dioxide. This is multiplied by the molar ratio of butane/CO2, (2/8) = 1/4, which gives the moles of butane required to produce the carbon dioxide.

Explanation:

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Makovka662 [10]

Answer:

1.51367e+10 inches

Explanation:

1 mile = 63360

63360 x 238900 = 15136704000

Hope this helped!

6 0
3 years ago
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
5 0
3 years ago
The feed to a batch process contains equimolar quantities of nitrogen and methane. write an expression for the kilograms of nitr
timama [110]
1) number of moles of N2 = n/2

2) Number of moles of CH4 = n/2

3) Total number of moles of the mixture = n/2 + n/2 = n

4) Kg of N2

mass in grams = number of moles * molar mass

molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol

=> mass of N2 in grams = (n/2) * 28 = 14n

mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg

Answer: mass of N2 in kg = 0.014n kg
3 0
3 years ago
If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f
I am Lyosha [343]

Answer:

0.032 L or 32 mL

Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

V2 = 2 L

Solve for V1 --> V1 = M2V2/M1

V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L

3 0
3 years ago
Describe the capillary action in acetone compared to water
lakkis [162]
Capillary action is defined as the ability of a liquid to go up a narrow space without the help or opposition of external forces. One of the most important factors affecting capillary action is the intermolecular forces within a substance. The higher the IMF, the greater the capillary action. The H-bonding in water gives it greater IMF than acetone, so water has greater capillary action.
6 0
3 years ago
Read 2 more answers
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