Which of these best describes the bonding between oxygen and hydrogen in the water molecule?

If 10.7 grams of NH4Cl is dissolved in enough water to make 800 mL of solution,what will be it's molarity

olga nikolaevna [1]

<h3><u>Answer;</u></h3>

Molarity = 0.25 M

<h3><u>Explanation;</u></h3>

Molarity is given by moles/Liter.

First we find moles:

Number of moles = Mass /molar mass

= (10.7g NH4Cl)/(53.5g/mol NH4Cl)

= 0.200 moles NH4Cl

Then we convert to liters:

= (800mL)*(1L/1000mL) = 0.800L

Therefore; molarity = 0.2moles/0.8L

= 0.25M

6 0

3 years ago

Mohr's salt is a pale green crystalline solid which is soluble in water. It is a 'double sulfate' which contains two cations, on

Veronika [31]

The gas is NH₃.
H₂ doesn't dissolve readily in water, SO₂ gives an acidic solution in water.

The solid residue is Fe(OH)₂.
FeSO₄ and Na₂SO₄ are soluble in water.

The answer is C.

3 0

3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

<u>The reaction </u>

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) <u>Limiting reagent</u>: $NO$

3) <u>Ammount of excess reagent</u>:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

How can mushrooms grow without harming trees?

son4ous [18]

It depends on the species because most mushrooms help by sharing the phosphorus with the tree

4 0

3 years ago

What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer

Zepler [3.9K]

$d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}$

8 0

3 years ago