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Mrrafil [7]
3 years ago
7

How many moles of carbon are in 7.87x10^7 carbon molecules?

Chemistry
1 answer:
zaharov [31]3 years ago
6 0
There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16
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A balloon filled with helium gas at 1.00 atm occupies 12.9 L. What volume would the balloon occupy in the upper atmosphere, at a
Taya2010 [7]

Answer:

67,9 L

Explanation:

Boyle's Law indicates that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of a gas, for a constant amount of gas we can write:

P1V1=P2V2

For the problem:

P1= 1 atm, V1= 12,9 L

P2=0,19 atm, V2=?

Therefore:

V2=P1V1/P2.................... V2=1 atm*12,9L/0,19 atm = 67,9 L

The balloon would occupy a volume of 67,9 L in the upper atmosphere.

5 0
3 years ago
Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution
Alla [95]

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

4 0
3 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
3 years ago
HELPPPPPPPPP ASAPPPP!!!!!
andreev551 [17]
Answer :
B

!!!!!!!!!!!!!!!
8 0
3 years ago
Ignore this<br>thanks (:
Vikki [24]
The answer is ignore this thanks
6 0
3 years ago
Read 2 more answers
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