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aleksklad [387]
2 years ago
6

Which of the following is a mass? Which is a Velocity?25kg2.5 meters/second ​

Chemistry
1 answer:
KATRIN_1 [288]2 years ago
7 0

Answer:

mass is 25

velocity is 2.5

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6Na(s) + N2(g) --> 2Na3N(s)
Ksivusya [100]

Answer:

 80.0 g Na and 20.0 g N2.

Explanation:

This means the limiting reactant determines the maximum mass of the product formed.

7 0
1 year ago
Relate pH values of 9.1, 1.2, and 5.7 to hydronium and hydroxyl ion concentration.
trapecia [35]

Answer:

9.1 = basic  1.2= very acidic  5.7= acidic

Explanation:

8 0
2 years ago
What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?
topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, R=0.08206\ L-atm/mol-K

Finding for V,

V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L

So, the volume of the gas is 34.55 L.

4 0
3 years ago
Oxygen gas is collected over water. The total pressure (the O2 pressure the water vapor pressure) is 736 torr. The temperature o
tigry1 [53]

Answer:

The value of the partial pressure of the oxygen  P_{O_{2} } =  690 torr

Explanation:

Total pressure of the mixture of gases = 736 torr

The partial pressure of water vapor = 46 torr

From the law of pressure we know that

Total pressure = The partial pressure of water vapor + The partial pressure of oxygen O_{2}

Put the values of pressures in above equation we get,

⇒ 736 = 46 + P_{O_{2} }

⇒ P_{O_{2} } = 736 - 46

⇒ P_{O_{2} } =  690 torr

This is the value of the partial pressure of the oxygen.

3 0
3 years ago
Compounds like CCl₂F₂ are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are n
natta225 [31]

Answer: Mass Of CFC that needs to evaporate for the freezing of water = 328.24 g

Explanation: Heat gained by the CFC = Heat lost by water

Heat lost by water = Heat required to take water's temperature to 0°c + Heat required to freeze water at 0°c

Heat required to take water's temperature from 33°c to 0°c = mCΔT

m = 201g, C = 4.18 J/(gK), ΔT = 33

mCΔT = 201 × 4.18 × 33 = 27725.94 J

Heat required to freeze water at 0°c = mL

m = 201g, L = 334 J/g

mL = 201 × 334 = 67134 J

Heat gained by CFC to vaporize = mH = 27725.94 + 67134 = 94859.94 J

H = 289 J/g, m = ?

m × 289 = 94859.9

m = 328.24 g

QED!!

7 0
3 years ago
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