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Vilka [71]
3 years ago
7

How many grams of NH4OH are present in 37.0 grams of 6.00 M NH4OH which has a density of 0.960 g/mL?

Chemistry
1 answer:
inn [45]3 years ago
6 0

Answer:

The mass of NH₄OH  present in the solution is 8.085 g

Explanation:

Given;

molarity of NH₄OH = 6 M

density of NH₄OH = 0.960 g/mL

mass of NH₄OH  = 37.0 grams

volume of  NH₄OH = mass /density

                                = 37/0.96

                                = 38.54 mL

molarity = moles of solute / Liters of solution

moles of solute  = molarity  x Liters of solution

moles of solute  = 6 x  38.54 x 10⁻³

moles of solute  = 0.231 moles

Reacting mass = number of moles x molecular mass

Molecular mass of NH₄OH = 35 g/mol

Reacting mass = 0.231 mol x 35 g/mol

Reacting mass = 8.085 g

Therefore, the mass of NH₄OH  present in the solution is 8.085 g

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How many sodium ions are in 4543.3 g of NaCl?
Stella [2.4K]

Answer:

4.68x10²⁵ ions of Na⁺

Explanation:

First of all, we dissociate the salt:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

An aqueous solution of NaCl dissociates in chlorides anions and sodium cations. Ratio is 1:1, per 1 mol of NaCl, we have 1 mol of Na⁺

We determine the moles of salt: 4543.3 g . 1mol / 58.45 g = 77.7 moles

77.7 moles are the amount of NaCl, therefore we have 77.7 moles of Na⁺.

We count the ions:

1 mol fo Na⁺ has 6.02x10²³ ions

77.7 moles of Na⁺ must have (77.7 . 6.02x10²³) / 1 = 4.68x10²⁵ ions of Na⁺

3 0
3 years ago
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An element's atomic number is 78. How many protons would an atom of this element have?
Oksanka [162]

Answer:

78

Explanation:

atomic number is number of protons

6 0
3 years ago
A solid and a liquid are shaken together in a test tube to produce a milky mixture that eventually separates into a clear liquid
Svet_ta [14]
It is insoluble. Because one of the substances are polar and the other is non-polar.

7 0
2 years ago
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How much calcium hypochlorite (65% strength) is needed to make 200 L of 2% hypochlorite solution? (Assume that a 1% solution is
g100num [7]

To solve this we use the equation,

 

M1V1 = M2V2

 

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

 

65 x V1 = 2 x 200 L

V1 = 6.15 L

5 0
3 years ago
A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3
ELEN [110]

Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

V_1=250ml

V_2=?

P_{total}=735 mmhg

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

P_{water}=42.2mmhg

T_1=35°C=35+273=308 K

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

P_{gas}=P_{total}-P_{water}

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}

V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1

V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1

Here P_2 \:and\: T_2 are standard pressure and temperature respectively.

we have

P_2=750mmhg \:and\: T_2=273K

Substituting value, we get

V_2=\frac{692.8}{750}*\frac{273}{308} *250

V_2= 819.51 ml

4 0
1 year ago
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