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Reptile [31]
3 years ago
15

What is the oxidation state of nitrogen in a nh4f molecule

Chemistry
1 answer:
joja [24]3 years ago
8 0

Answer:

+3

Explanation:

NH₄F

=> N + 4H + F = 0

=> N + 4(+1) + 1(-7) = 0

=> N = 7 - 4 = +3

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Which of the following terms is defined by the measure of mass per unit of volume?
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Density is defined by the above term
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Crude oil is ________. a. composed of less than ten different hydrocarbon molecules b. formed in a wide range of temperature and
Arisa [49]

Answer:

C

Explanation:

A - Crude oil is composed for hundreds of hydrocarbon, not less than ten.

B - Is formed in specific conditions of temperature and pressure

C - It's fractionated to form gasoline, lubricants, CH4, plastics and many other products made of hydrocarbon.

D - We have crude oil located more than 4000 yards below the surface in countries like Brazil

E - The crude oil is very thick and don't have an specific usage, so we need to refine it.

3 0
3 years ago
A gas has a volume of 5.00 L at 0°C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to ea
Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

C = -268.37°C

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C  (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

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Determining what it is you want to know:
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