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Varvara68 [4.7K]
3 years ago
5

It has been raining all day and the temperature is expected to freeze tonight. Mrs. Gussman treats her front steps with sodium c

hloride (NaCl). This results in a water and salt solution with a freezing point that is lower than pure water. If 850 grams of sodium chloride is added to 6.10 kilograms of water, what will be the freezing point for this solution? The freezing-point depression constant for water is Kf = –1.86°C/m. In solving this problem, consider that sodium breaks into Na+ and Cl– ions, doubling the molar concentration of the salt.
Chemistry
2 answers:
Komok [63]3 years ago
6 0

Answer:

850Grams of sodium chloride+6.10kilograms of water.

Explanation:It has been raining all day and the temperature is expected to freeze tonight. Mrs. Gussman treats her front steps with sodium chloride (NaCl). This results in a water and salt solution with a freezing point that is lower than pure water. <u>If 850 grams of sodium chloride is added to 6.10 kilograms of water, what will be the freezing point for this solution?</u> The freezing-point depression constant for water is Kf = –1.86°C/m.<u> In solving this problem, consider that sodium breaks into Na+ and Cl– ions, doubling the molar concentration of the salt.</u>

<u />

The underlined words where your help to find the awnser.

VMariaS [17]3 years ago
3 0

Answer:

It has been raining all day and the temperature is expected to freeze tonight. Mrs. Gussman treats her front steps with sodium chloride (NaCl). This results in a water and salt solution with a freezing point that is lower than pure water. If 850 grams of sodium chloride is added to 6.10 kilograms of water, what will be the freezing point for this solution? The freezing-point depression constant for water is Kf = –1.86°C/m. In solving this problem, consider that sodium breaks into Na+ and Cl– ions, doubling the molar concentration of the salt.It has been raining all day and the temperature is expected to freeze tonight. Mrs. Gussman treats her front steps with sodium chloride (NaCl). This results in a water and salt solution with a freezing point that is lower than pure water. If 850 grams of sodium chloride is added to 6.10 kilograms of water, what will be the freezing point for this solution? The freezing-point depression constant for water is Kf = –1.86°C/m. In solving this problem, consider that sodium breaks into Na+ and Cl– ions, doubling the molar concentration of the salt.

Explanation:

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How many moles of N2O5 are needed<br> to produce 7.90 g of NO2?
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Answer:

0.085 moles of  N₂O₅ are needed

Explanation:

Given data:

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Moles of N₂O₅ needed = ?

Solution:

2N₂O₅       →     4NO₂  + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

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now we will compare the moles of NO₂   with N₂O₅.

                NO₂          :       N₂O₅

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Answer:

0.7457 g is the mass of the helium gas.

Explanation:

Given:  

Pressure = 3.04 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25.0 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K  

<u>⇒n = 0.1863 moles</u>

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The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.1863\ mole= \frac{Mass}{4.0026\ g/mol}

Mass_{He}= 0.7457\ g

<u>0.7457 g is the mass of the helium gas. </u>

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