What is the mass in pounds of 1.000 quart of mercury

If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the mola

rodikova [14]

Volume Ba(OH)2 = 23.4 mL in liters :

23.4 / 1000 => 0.0234 L

Molarity Ba(OH)2 = 0.65 M

Volume HNO3 = 42.5 mL in liters:

42.5 / 1000 => 0.0425 L

number of moles Ba(OH)2 :

n = M x V

n = 0.65 x 0.0234

n = 0.01521 moles of Ba(OH)2

Mole ratio :

Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O

1 mole Ba(OH)2 ---------------- 2 moles HNO3
0.01521 moles ----------------- moles HNO3

moles HNO3 = 0.01521 x 2 / 1

moles HNO3 = 0.03042 / 1

= 0.03042 moles HNO3

Therefore:

M ( HNO3 ) = n / volume ( HNO3 )

M ( HNO3 ) = 0.03042 / 0.0425

M ( HNO3 ) = 0.715 M

5 0

3 years ago

What subatomic particle(s) is on the outside of the atom?

asambeis [7]

Answer:

electrons

Explanation:

protons and neutrons are found in the nucleus while electrons are outside the nucleus

4 0

4 years ago

Is HF the most polar among covalent bonds?

Sladkaya [172]

Answer:

No.

Explanation:

HI is more polar than HF

3 0

2 years ago

Given the following equation: 4 NH3 (g)5 O2 (g) >4 NO (g) + 6 H20 (I) How many moles of NH3 is required to react with 25.7 gr

Sergeu [11.5K]

Answer:

0.6425 moles of $NH_3$ is required to react with 25.7 grams of $O_2$ .

Explanation:

mas of oxygen gas = 25.7 g

moles of oxygengas = $\frac{25.7 g}{32 g/mol}=0.8031 mol$

$4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)$

According to reaction given above, 5 moles of oxygen gas reacts with 4 moles of ammonia gas.

Then 0.8031 moles of oxygen gas will react with :

$\frac{4}{5}\times 0.8031 mol=0.6425 mol$ of ammonia gas

0.6425 moles of $NH_3$ is required to react with 25.7 grams of $O_2$ .

5 0

3 years ago

How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl

kogti [31]

Answer:

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:

Mg: 1 mole
HCl: 2 moles
MgCl₂: 1 mole
H₂: 1 mole

Being:

Mg: 24. 31 g/mole
H: 1 g/mole
Cl: 35.45 g/mole

the molar mass of the compounds participating in the reaction is:

Mg: 24.31 g/mole
HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
H₂: 2*1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

Mg: 1 mole* 24.31 g/mole= 24.31 g
HCl: 2 moles* 36.45 g/mole= 72.9 g
MgCl₂: 1 mole* 95.21 g/mole= 95.21 g
H₂: 1 mole* 2 g/mole= 2 g

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

$moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}$

moles of H₂= 0.134

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

7 0

4 years ago