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3 years ago

A sample of 1.00 moles of oxygen at 50°C and 98.6 kPa, occupies what volume?

Chemistry

1 answer:

enyata [817]3 years ago

5 0

Answer:

27.22 dm³

Explanation:

Given parameters:

number of moles = 1 mole

temperature= 50°C, in K gives 50+ 273 = 323K

Pressure= 98.6kpa in ATM, gives 0.973 ATM

Solution:

Since the unknown is the volume of gas, applying the ideal gas law will be appropriate in solving this problem.

The ideal gas law is mathematically expressed as,

Pv=nRT

where P is the pressure of the gas

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

Input the parameters and solve for V,

0.973 x V = 1 x 0.082 x 323

V= 27.22 dm³

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If you had 1.50 g of ZnSO4.7H20, zinc sulfate septahydrate, how many grams of water will there be?

MariettaO [177]

Answer:

There will be 1.50 gram of water

5 0

3 years ago

Consider the following reaction: COCl2(g) ⇌ CO(g) + Cl2(g) A reaction mixture initially contains 1.6 M COCl2. Determine the equi

professor190 [17]

Answer:

The equilibrium concentration of CO is 0.0361 M

Explanation:

Step 1: Data given

Kc = 8.33 *10^-4

Molarity of COCl2 = 1.6 M

Step 2: The balanced equation:

COCl2(g) ⇌ CO(g) + Cl2(g)

Step 3: Calculate final concentrations

The initial concentration of COCl2 = 1.6M

The initial concentration of CO and Cl2 = 0M

There will react xM of COCl2

Since the mole ratio is 1:1

The final concentration of CO and Cl2 will be X M

The final concentration of COCl2 will be (1.6 -X)M

Step 4: Define Kc

Kc= [CO] *[Cl2] / [COCl2] = 8.33*10^-4

Kc = X*X / 1.6-X = 8.33 * 10^-4

8.33 * 10^-4 = X² /(1.6-X)

8.33 * 10^-4 *(1.6 -X) = X²

0.0013328 - 8.33*10^-4 X = X²

X² + 8.33*10^-4 X - 0.0013328= 0

X = 0.0361 M = [CO] = [Cl2]

[COCl2] = 1.6 - 0.0361 = 1.5639 M

To control this we can calculate the Kc

(0.0361*0.0361)/1.5639 = 0.000833

5 0

3 years ago

A 55.9 g sample of water at 99.2 °c is placed in a constant pressure calorimeter. then, 23.9 g of zinc metal at 21.1 °c is added

Veronika [31]

<span>Heat released by the water: mass * specific heat * ΔT

Heat released by the water = 55.9 g * 1 cal / (°C g) * (99.2 - 96.9 )°C

Heat absorbed by the zinc metal = mass * specifi heat * ΔT

Heat absorbed by the zinc metal = 23.9 g * Cp * (96.9 - 21.1) °C

Heat absorbed by zinc metal = heat released by water

=> 23.9 * Cp * 75.8 = 55.9 * 1 * 2.3

=> Cp = 128.57 / 1811.62 = 0.0071 cal / (g°C)

Answer: 0.0071 cal / (g°C)
</span>

4 0

3 years ago

What 4 elements whose isotopes have a half-life of more than one year?

Oliga [24]

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6 0

3 years ago

Which one is it ASAP

katrin2010 [14]

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8 0

3 years ago