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storchak [24]
3 years ago
9

What is the concentration of a solution in which 10.0 g of AgNO, is dissolved in 450 mL of solution?

Chemistry
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer:

M=0.15

Explanation:

138 g AgNO -> 1 mol AgNO

10 g AgNO   -> x

x= (10 g AgNO * 1 mol AgNO)/138 g        x=0.07 mol AgNO

450 mL=0.45 L

M= mol solute/L solution

M= 0.07 mol AgNO/0.45L

M=0.15

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find the the number  of moles

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= 0.738/3500 x1000  = 0.21 M
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What volume would 3.01•1023 molecules of oxygen gas occupy at STP?
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What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer
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8 0
3 years ago
Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10
PtichkaEL [24]

Answer:

K.E.=1.97\times 10^{-21}\ J

Explanation:

Given that:-

Pressure = 6.9\times 10^5\ Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = \frac {1}{101325} P (atm)

Given the value of pressure = 43,836 Pa

So,  

6.9\times 10^5\ Pa = \frac{6.9\times 10^5}{101325} atm

Pressure = 6.80977 atm

Volume = 2.3\times 10^{-3}\ m^3 = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

K.E.=\frac{3}{2}\times K\times T

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

T is the temperature

So, K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J

K.E.=1.97\times 10^{-21}\ J

3 0
3 years ago
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