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Sonbull [250]
3 years ago
6

A) What mass in grams of H20 is needed to react completely with 40.0 g of

Chemistry
1 answer:
Step2247 [10]3 years ago
7 0

Answer:

m_{H_2O}=9.24gH_2O

Explanation:

Hello there!

In this case, since there is a 2:2 mole ratio between sodium peroxide and water according to the given reaction, it is possible to apply the following stoichiometric setup for the calculation of the required mass of water:

m_{H_2O}=40.0gNa_O_2*\frac{1molNa_O_2}{78gNa_O_2}*\frac{2molH_2O}{2molNa_O_2}  *\frac{18.02gH_2O}{1molH_2O} \\\\m_{H_2O}=9.24gH_2O

Best regards!

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Answer:

Conductivity meter

Explanation:

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A pH paper is used to determine the pH of a solution. This is done by dipping part of the paper into a solution of interest and watching the color change. The pH paper comes in a color-coded scale indicating the pH that something has when the paper turns a certain color.

An indicator is an organic compound that changes its colour depending on the pH of the solution.

Since neutralization reaction can only be monitored by monitoring the pH of the solution, a conductivity meter cannot be used to monitor the progress of a neutralization reaction since it does not monitor the change in pH of the system under study.

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What happens when magnesium reacts with dilute acid?
iogann1982 [59]

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When is work considered to be work?<br> (this is about force,work and energy)
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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
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