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ad-work [718]
3 years ago
6

Sodium chloride reacts with 0.10 moles of aluminum nitrate in water soultion. How many grams of sodium chloride are required?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

The answer to your question is 0.3 moles of NaCl

Explanation:

Data

NaCl = ?

Al(NO₃)₃ = 0.1 moles

Balanced reaction

                         Al(NO₃)₃  + 3NaCl    ⇒   3NaNO₃   +   AlCl₃

Now, use proportions to solve the problem

                       1 mol of Al(NO₃)₃  ------------ 3 moles of NaCl

                     0.1 moles of Al(NO₃)₃  --------  x

                          x = (0.1 x 3) / 1

                          x = 0.3 moles of NaCl

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Answer:

D. water

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7 0
3 years ago
A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution
Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

<em></em>

I hope it helps!

7 0
3 years ago
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6 0
3 years ago
Cl2+2KI→2KCl + I2 is an example of which type of reaction?
Gnoma [55]

Answer:

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Explanation:

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3 years ago
When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat
Angelina_Jolie [31]

Answer:

The question has some details missing. here are the details ; Given the following ;  

1. 43.2 g of tablet with 20 cm3 of space

2. 5 cm3 of tablets weighs 10.8 g

3. 5 g of balsa wood with density 0.16 g/cm3

4. 150 g of iron. With density 79g/cm 3

5. 32 cm3 sample of gold with density 19.3 g/cm3

6. 18 ml of cooking oil with density 0.92 g/ml

Explanation:

<u>Appropriate for calculating mass</u>

32 cm3 sample of gold with density 19.3 g/cm3

18 ml of cooking oil with density 0.92 g/ml

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<u>Appropriate for calculating density</u>

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3 years ago
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