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tensa zangetsu [6.8K]
3 years ago
6

A. At STP, what is the volume of 708 mol of nitrogen gas? 708 mol = 708 mol X L B. A sample of hydrogen gas occupies 14.1 L at S

TP. Hov many moles of the gas are present? 14.1 L = 14.1 L X 11 mol​
Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

A. 15859.2 L or 15900 L

B. 0.629 mol

Explanation:

At STP, one mole is equal to approximately 22.4 L

L or mL is volume, so you are attempting to solve for L or mL.

A.

708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)

B.

(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.

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How many atoms of Al(OH)3 in a 72 gram sample? please show work​
Lostsunrise [7]

Answer:

5.559*10^(23) atoms

Explanation:

Molar Mass of Al(OH)3 = 78 g/mol

(72g/78g/mol)*(6.022*10^(23)atoms/mol) = 5.559*10^(23) atoms in 72 grams of Al(OH)3

4 0
4 years ago
When 0.873 grams of a protein were dissolved in 48.6 mL of solution at 15.6 degrees C, the osmotic pressure was found to be 0.06
Stels [109]

Answer:

697 g / mol

Explanation:

The osmotic pressure of a protein ( non electrolyte ) is given by:

π V= nRT where  V is the volume, n is the number of moles, R is the gas constant ( 0.08206 L·atm/Kmol ), and T is the temperature (K).

n= mass/ MW protein ⇒ MW protein = mass / n

Thus,

π V = ( mass/ MW ) RT

MW = mass x R xT/ ( π V )

mass = 0.873 g

R = 0.08206 L·atm/K·mol

T = ( 15.6 + 273 ) K= 288.6 K

π  = 0.061 atm

V = 48.6 mL = 48.6 mL x ( 1 L/ 1000 mL ) = 0.0486 L

We just need to plug our values into the aqbove equation for MW:

MW = 0.873 g x 0.08205 L· atm /K·mol x 288.6 K / ( 0.061 atm x0.0486 L )

       = 697 g/mol

6 0
4 years ago
What do acids do in solution?
Rasek [7]

Answer:

C.

Explanation:

The increase the [H+] (concentration of H+ ions)

However, they decrease the pH if acids are added to any solution. Since they have more [H+] ions, they increase their conc. in the solution

5 0
3 years ago
at a pressure of 405 kpa and a temperature of 200k the volume of a gas is 6.00cm^3. At what pressure will the new temperature an
Bas_tet [7]

idek what that is but i have the question on my homework rn

6 0
3 years ago
25.0g of iron is heated to 100.0 and then placed in 50.0 g of water in a insulated calorimeter. the initial temperature of the w
Vinvika [58]

Answer:

Approximately 41.2\; {\rm ^{\circ} C}.

Explanation:

Let t\; {\rm ^{\circ} C} be the final temperature of the water and the iron.

Temperature of the water would be increase by (t - 38.00)\; {\rm ^{\circ} C}.

Temperature of the iron would be reduced by (100.0 - t)\; {\rm ^{\circ} C}.

Let c denote the specific heat of each material. Let m denote the mass of the material. For a temperature change of \Delta t, the energy change involved would be:

Q = c\, m \, \Delta t.

The energy that the water need to absorb would be:

\begin{aligned}& Q(\text{water, absorbed}) \\ =\; & c(\text{water}) \, m(\text{water})\, \Delta t (\text{water}) \\ =\; & 4.181\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 50\; {\rm g} \times (t - 38.00)\; {\rm ^{\circ} C} \\ =\; & (209.05\, t - 7943.9)\; {\rm J} \end{aligned}.

The energy that the iron would need to release would be:

\begin{aligned}& Q(\text{iron, released}) \\ =\; & c(\text{iron}) \, m(\text{iron})\, \Delta t (\text{iron}) \\ =\; & 0.45\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 25.0\; {\rm g} \times (100.0 - t)\; {\rm ^{\circ} C} \\ =\; & (1125 - 11.25 \, t)\; {\rm J} \end{aligned}.

Since this calorimeter is insulated, the energy that the iron had released would be equal to the energy that the water had absorbed:

Q(\text{water, absorbed}) = Q(\text{iron, released}).

209.05\, t - 7943.9 = 1125 - 11.25\, t.

t \approx 41.2.

Thus, the final temperature of the water and the iron would be approximately 41.2\; {\rm ^{\circ} C}.

5 0
3 years ago
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