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3 years ago

The most dangerous form of skin cancer which is often caused by over exposure to ultraviolet radiation is what?

1 answer:

8 0

UV exposure that leads to sunburn is proven in developing melanoma, the most dangerous of the three most common types of skin cancer.

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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

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

Wave particle duality is most apparent in analyzing the motion of.

DIA [1.3K]

Answer:

Wave particle duality is most apparent in analyzing the motion of. an electron. A photon of which electromagnetic radiation has the most energy.

5 0

2 years ago

16. A solution is made by dissolving 25 g of NaCl in enough water to make 1.0 L of solution. Assume the density of the solution

nata0808 [166]

Answer:

The molarity will be "0.43 M" and molality will be "0.43 m".

Explanation:

Given that:

mass of NaCl,

= 25 g

Volume of solution,

= 1 L

Density of solution,

= 1 gm/cm³

Now,

The weight of solvent will be:

= $Volume\times density$

= $1\times 1$

= $1 \ kg$

The mole of NaCl will be:

= $\frac{mass}{Molar \ mass}$

= $\frac{25}{58.44}$

= $0.43$

hence,

The molarity will be:

= $\frac{number \ of \ mole}{Volume \ of \ solution}$

= $\frac{0.43}{1}$

= $0.43 \ M$

The molality will be:

= $\frac{mole \ of \ solute}{Weight \ of \ solvent}$

= $\frac{0.43}{1}$

= $0.43 \ m$

4 0

3 years ago

In a gas grill, 29 lbs propane C3H8 are

dimulka [17.4K]

Answer : The mass of combustion products formed are 134 lbs.

Explanation :

The balanced chemical reaction will be:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Given :

Mass of $C_3H_8$ = 29 lbs = 13154.2 g

conversion used : 1 lbs = 453.592 g

Molar mass of $C_3H_8$ = 44 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles$

Now we have to calculate the moles of $CO_2$ and $H_2O$ .

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_3H_8$ react to give 3 moles of $CO_2$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 3=896.7$ moles of $CO_2$

and,

As, 1 mole of $C_3H_8$ react to give 4 moles of $H_2O$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 4=1195.6$ moles of $H_2O$

Now we have to calculate the mass of $CO_2$ and $H_2O$ .

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2$

$\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs$

and,

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O$

$\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs$

The total mass of products = Mass of $CO_2$ + Mass of $H_2O$

The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs

Therefore, the mass of combustion products formed are 134 lbs.

6 0

3 years ago