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3 years ago

What is the percent by mass of Fluorine in Nitrogen triflouride NF3?

Chemistry

1 answer:

Kipish [7]3 years ago

3 0

Answer:

The answer to your question is 80.3%

Explanation:

Data

Percent by mass of F

molecules NF₃

Process

1.- Calculate the molar mass of nitrogen trifluoride

molar mass = (1 x 14) + (19 x 3)

= 14 + 57

= 71 g

2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.

71 g of NF₃ ------------------ 100%

57 g of F ------------------- x

x = (57 x 100)/71

x = 5700 / 71

x = 80.3%

3.- Conclusion

Fluorine is 80.3% by mass of the molecule NF₃

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The chemical name of copper ii tetra sulphate pentahydrate

Anastasy [175]

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3 0

2 years ago

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO

zhenek [66]

Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x $\frac{3}{2}$
= 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg

Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%

7 0

3 years ago

Formula for water what does the lack of a subscript after those O indicate

photoshop1234 [79]

It indicates that there is only one oxygen molecule

3 0

2 years ago

When the volume of a gas is

Cerrena [4.2K]

Answer:

THE NEW VOLUME OF THE GAS IS 406 mL WHEN THE TEMPERATURE CHANGES FROM 765 K TO 315 K.

Explanation:

When the temperature changes from 765 K to 315K, the volume has changed from 986 mL to?

V1 = 986 mL = 0.986 L

T1 = 765 K

T2 = 315 K

V2 = unknown

Using Charles' equation of gas laws;

V1 / T1 = V2 / T2

Making V2 the subject of the formula:

V2 = V1 T2 / T1

V2 = 0.986 * 315 / 765

V2 = 0.406 L

V2 = 406 mL

So therefore, the volume of a gas changes from 986 mL to 406 mL as a result of a change in temperature from 765 K to 315 K.

7 0

3 years ago

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an

Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0

3 years ago