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ruslelena [56]
3 years ago
13

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

Chemistry
1 answer:
koban [17]3 years ago
6 0

Answer: A: 10

Explanation:

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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} )

<em>We see that Liters and Liters cancel out, so we simply multiply.</em>

847 mL

7 0
3 years ago
35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--&gt; 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

|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|

|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|

<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|

|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>

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
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