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statuscvo [17]
3 years ago
14

Consider the unbalanced equation for the combustion of hexane:

Chemistry
1 answer:
krek1111 [17]3 years ago
8 0
Ratio of C6H14 to O2 
2:19 
2/19=7.9/x
solve for x and youre done!! 
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In the presence of a base, blue litmus paper will .<br>O turn purple<br>stay blue<br>to turn red​
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In the presence of a base, blue litmus paper will turn red........

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
How did the scientific method change the way scientists worked beginning in
earnstyle [38]

Answer:

A). It encouraged them to rely on observation and experimentation to support their conclusions.

Explanation:

3 0
3 years ago
How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?<br><br><br> Read bottom comment
Genrish500 [490]

Answer:

The following relationship makes this possible: 1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.

Explanation:

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How many moles are in 25 g of NaCl
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4.5098 moles i believe'
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