What sound frequency would a human ear not be able to detect?

All the substances listed below are fertilizers that contribute nitrogen to the soil.

kvv77 [185]

Answer:

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

Explanation:

Percentage of element in compound :

$=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100$

(a) Urea, $(NH_2)_2CO$

Molar mass of urea = 60 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

$N\%=\frac{2\times 14 g/mol}{60 g/mol}\times 100=46.67\%$

(b) Ammonium nitrate, $NH_4NO_3$

Molar mass of ammonium nitrate = 80 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

$N\%=\frac{2\times 14 g/mol}{80 g/mol}\times 100=35.00\%$

(c) Nitric oxide, NO

Molar mass of nitric oxide = 30 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

$N\%=\frac{1\times 14 g/mol}{30 g/mol}\times 100=46.67\%$

(d) Ammonia, $NH_3$

Molar mass of ammona = 17 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

$N\%=\frac{1\times 14 g/mol}{17 g/mol}\times 100=82.35\%$

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

4 0

4 years ago

How many milliliters (mL) are in 0.847 liters (L)

alex41 [277]

Answer:

847 mL

Explanation:

Step 1: Find conversion

1 L = 1000 mL

Step 2: Use Dimensional Analysis

$0.847 \hspace{2} L(\frac{1000 \hspace{2} mL}{1 \hspace{2} L} )$

We see that Liters and Liters cancel out, so we simply multiply.

847 mL

7 0

3 years ago

35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--> 2NH3

Misha Larkins [42]

Explanation:

Moles of N2 = 35.0g / (28g/mol) = 1.25mol

Moles of H2 = 60.0g / (2g/mol) = 30.0mol

Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.

Moles of NH3 = 1.25mol * 2 = 2.50mol.

Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.

30.0mol - 1.25mol * 3 = 26.25mol.

Excess mass of H2

= 26.25mol * (2g/mol) = 52.5g.

6 0

3 years ago

Who at the top of the food web

yKpoI14uk [10]

Humans but if that isn’t the answer there it is probably talking about a certain animal group

7 0

3 years ago

Read 2 more answers

Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

3 years ago