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2 years ago

Which weather condition is NOT associated with warm fronts??

Chemistry

1 answer:

slavikrds [6]2 years ago

5 0

Answer:

A Winter storm usually does not come with a warm fronthey warm front.

Explanation:

When the front passes over an area it means to change in the weather. Many. Many friends cause weather events such as rain thunderstorms gusty winds tornadoes and hurricanes.

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Convert 22.4 kg/L to kg/mL

liberstina [14]

1 kg/L -------------- 0.001 kg/mL
22.4 kg/L --------- ??

22.4 x 0.001 / 1 => 0.0224 kg/mL

7 0

3 years ago

A tank at is filled with of sulfur tetrafluoride gas and of sulfur hexafluoride gas. You can assume both gases behave as ideal g

dangina [55]

The question is incomplete, the complete question is:

A 7.00 L tank at $21.4^oC$ is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

Answer: The mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

For sulfur hexafluoride:

Given mass of sulfur hexafluoride = 5.43 g

Molar mass of sulfur hexafluoride = 146.06 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol$

For sulfur tetrafluoride:

Given mass of sulfur tetrafluoride = 14.2 g

Molar mass of sulfur tetrafluoride = 108.07 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfur tetrafluoride }=\frac{14.2g}{108.07g/mol}=0.1314mol$

Total moles of gas in the tank = [0.0372+ 0.1314] mol = 0.1686 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .....(2)

where n is the number of moles

Putting values in equation 2, we get:

$\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221$

$\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779$

Hence, the mole fraction of sulfur hexafluoride is 0.221 and that of sulfur tetrafluoride is 0.779

7 0

3 years ago

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

$\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g$

Therefore, the mass of $NaN_3$ required is, 166.4 grams.

3 0

3 years ago

What is the proper name for Zn(CIO)2?

Sergio [31]

Answer:

ZINC CHLORATE

Explanation:

Zinc chloride is the name of chemical compounds with the formula ZnCl2 and its hydrates.

3 0

3 years ago

Read 2 more answers

In a furnace zinc is a gas but lead is a liquid. Suggest why?

Tems11 [23]

I believe it is because, Lead has a higher boiling point than zinc. by the time Lead melts and becomes a liquid, the temperature is way higher than the boiling point of Zinc therefore, Zinc becomes a gas.

6 0

3 years ago