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olasank [31]
2 years ago
7

Calculate the grams of nitrogen in 125 g of NH4NO3

Chemistry
1 answer:
Marta_Voda [28]2 years ago
5 0

Answer:

43.75 g of Nitrogen

Explanation:

We'll begin by calculating the mass of 1 mole of NH₄NO₃. This can be obtained as follow:

Mole of NH₄NO₃ = 1 mole

Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)

= 14 + 4 + 14 + 48

= 80 g/mol

Mass of NH₄NO₃ =?

Mass = mole × molar mass

Mass of NH₄NO₃ = 1 × 80 = 80 g

Next, we shall determine the mass of N in 1 mole of NH₄NO₃.

Mass of N in NH₄NO₃ = 2N

= 2 × 14

= 28 g

Thus,

80 g of NH₄NO₃ contains 28 g of N.

Finally, we shall determine the mass of N in 125 g of NH₄NO₃. This can be obtained as follow:

80 g of NH₄NO₃ contains 28 g of N.

Therefore, 125 g of NH₄NO₃ will contain = (125 × 28) / 80 = 43.75 g of N.

Thus, 125 g of NH₄NO₃ contains 43.75 g of Nitrogen

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mariarad [96]
The answer is B

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3 0
3 years ago
I really need the answers please!
krek1111 [17]

1) D = 13.6 g / mL

2)ethyl alcohol weighs 158g

3)ρ _copper = 8.9 g cm^{3}

Explanation:

1)

D = m / V

=306.0 g / 22.5 mL

D= 13.6 g / mL

2)

density = mass / volume

mass = density × volume  

=0.789g /ml × 200.0 ml

M=158g

Ethyl alcohol weighs 158g

3)

ρ  (density) = Mass  / Volume

ρ _copper = 1896 g / 8.4cm × 5.5cm × 4.6cm

= 1896g / 212.5 cm^{3}

ρ _copper=8.9 g cm^{3}

4 0
2 years ago
Diethyl ether is produced from ethanol according to the following equation:
juin [17]
The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of  CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3 

% yield = .775 = actual yield / 264.66


actual yield = 205.11 g CH3CH2OCH2CH3
6 0
3 years ago
Calculate the grams of CaCl2 necessary to make a 0.15Msolution.
xxTIMURxx [149]

Answer: mass m = M·c·V

Explanation: M(CaCl2) = 110.98 g/mol, c= 0.15 mol/l,

n=m/M= cV, volume of Solution is not mentioned

3 0
2 years ago
How many milliliters of sterile water for injection should be added to a vial containing 5 mg/mL of a drug to prepare a solution
jonny [76]

Answer:

7mL of sterile water is the initial amount of the concentrated solution is 3mL

Explanation:

In this problem, the vial must be <em>diluted </em>from 5mg/mL to 1.5mg/mL, that means the solution must be diluted:

5mg/mL / 1.5mg/mL = 3.33 times

If the initial amount of the drug in the vial is 3mL, the final volume must be:

10mL

That means the volume of water that should be added is:

10mL - 3mL:

<h3>7mL of sterile water is the initial amount of the concentrated solution is 3mL</h3>
4 0
3 years ago
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