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2 years ago

How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?

1 answer:

4 0

When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed

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If a certain gas occupies a volume of 13 L when the applied pressure is 6.5 atm , find the pressure when the gas occupies a volu

Sav [38]

We should apply Boyle's Law here given initial pressure, initial volume and final volume.

P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)

Solve for P2 on your calculator and that should get you to the answer.

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2 years ago

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In a container containing co, h2, and o2, what is the mole fraction of co if the h2 mole fraction is 0.22 and the o2 mole fracti

almond37 [142]

Thank you for posting your question here at brainly. The mole fraction of co if the h2 mole fraction is 0.22 and the o2 mole fraction is 0.58 is 0.20, below is the solution:

mole fraction CO + mole fraction H2 + mole fraction O2 = 1

mole fraction CO = 1 - ( 0.22 + 0.58)=0.20

6 0

2 years ago

24 how many moles are in 2.04 × 1024 molecules of h2o?

Sergeeva-Olga [200]

The answer is 3.39 mol.

Avogadro's number is the number of molecules in 1 mol of substance.
6.02 × 10²³ molecules per 1 mol.
2.04 × 10²⁴ molecules per x.

6.02 × 10²³ molecules : 1 mol = 2.04 × 10²⁴ molecules : x
x = 2.04 × 10²⁴ molecules * 1 mol : 6.02 × 10²³ molecules
x = 2.04/ 6.02 × 10²⁴⁻²³ mol
x = 0.339 × 10 mol
x = 3.39 mol

8 0

2 years ago

At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2

s344n2d4d5 [400]

Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - Henry's law constant.
c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

4 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

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2 years ago

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