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Alenkinab [10]
2 years ago
8

A stock solution of ammonia in water is 28 wt% NH3. The Formal Weight (FW) of NH3 is 17.03 g/mol). The solution density at room

temperature is 0.90 g/mL. What is the molarity of the solution in units of mol/L
Chemistry
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

[NH₃] = 14.7 mol/L

Explanation:

28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.

Let's determine the amount of ammonia:

28 g . 1 mol / 17.03g = 1.64 moles of NH₃

You need to consider that, when you have density's data it is always referred to solution:

Mass of solution is 100 g, let's find out the volume

0.90 g/mL = 100 g /V

V = 100 g / 0.90mL/g → 111.1 mL

We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L

mol/L = 1.64 mol/0.1111L → 14.7 M

mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution

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Explanation:

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

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s = 1 orbital

p = 3 orbitals

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f = 7 orbitals

For n = 4

l = 0 to (n-1) = 0 to 3 = (4s , 4p , 4d , 4f)

Number of subshells = 4

Number of orbitals =         1 + 3 + 5 + 7  = 16

The maximum number of electrons the n = 4 shell can contain:

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16\times 2=32

32 is the maximum number of electrons the n = 4 shell can contain

6 0
3 years ago
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What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?
Aleksandr [31]

Answer:

\boxed {\boxed {\sf 3.3 \ liters}}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

  • molarity= 1.2 mol/L
  • moles of solute= 4.0 mol
  • liters of solution =x

Substitute the values into the formula.

1.2 \ mol/L = \frac{4.0 \ mol}{x}

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}

4.0 \ mol *1=1.2 \ mol/L *x

4.0 \ mol = 1.2 \ mol/L *x

x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L

\frac{4.0 \ mol}{1.2 \ mol/L} = \frac{1.2 \ mol/L *x}{1.2 \ mol/L}

\frac{4.0 \ mol}{1.2 \ mol/L}=x

The units of moles (mol) will cancel.

\frac{4.0 }{1.2 } \ L =x

3.33333333 \ L=x

The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.

The 3 in the hundredth place tells us to leave the 3 in the tenths place.

3.3 \ L\approx x

Approximately  <u>3.3 liters of solution</u> are needed to make a 1.2 M solution with 4.0 moles of sodium chloride.

7 0
3 years ago
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Consider the oxidation of sodium metal to sodium oxide described by the balanced equation:
mixas84 [53]

Answer:

1116 g.

Explanation:

The balanced equation for the reaction is given below:

4Na + O₂ —> 2Na₂O

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of Na₂O.

Next, we shall determine the theoretical yield of Na₂O. This can be obtained as follow:

From the balanced equation above,

1 mole of O₂ reacted to produce 2 moles of Na₂O.

Therefore, 9 moles of O₂ will react to produce = 9 × 2 = 18 moles of Na₂O.

Finally, we shall determine the mass in 18 moles of Na₂O. This can be obtained as follow:

Mole of Na₂O = 18 moles

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Mass of Na₂O =?

Mass = mole × molar mass

Mass of Na₂O = 18 × 62

Mass of Na₂O = 1116 g

Thus, the theoretical yield of Na₂O is 1116 g.

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