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valkas [14]
3 years ago
10

How many formula units make up 10.2 g of magnesium chloride (MgCl2)?

Chemistry
1 answer:
beks73 [17]3 years ago
6 0

Answer:

6.46×10²² formula units

Explanation:

From the question given above, the following data were obtained:

Mass of MgCl₂ = 10.2 g

Number of formula units =?

From Avogadro's hypothesis,

1 mole of MgCl₂ = 6.02×10²³ formula units

But,

1 mole of MgCl₂ = 24 + (35.5×2) = 24 + 71 = 95 g

Thus, we can say:

95 g of MgCl₂ = 6.02×10²³ formula units

Finally, we shall determine the formula units in 10.2 g MgCl₂. This can be obtained as follow:

95 g of MgCl₂ = 6.02×10²³ formula units

Therefore,

10.2 g of MgCl₂ = (10.2 × 6.02×10²³) / 95

10.2 g of MgCl₂ = 6.46×10²² formula units

Thus, 10.2 g of MgCl₂ contains 6.46×10²² formula units

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How many moles of water are in a beaker with 50 mL?
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Answer:

Number of moles = 2.8 mol

Explanation:

Given data:

Number of moles of water = ?

Volume of water = 50 mL

Density of water = 1.00 g/cm³

Solution:

1 cm³ =  1 mL

Density = mass/ volume

1.00 g/mL = mass/ 50 mL

Mass = 1.00 g/mL× 50 mL

Mass = 50 g

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 50 g / 18 g/mol

Number of moles = 2.8 mol

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3 years ago
Solution A has a pH of 10, and solution B has a pH of 1. Which statement best describes these solutions?
Free_Kalibri [48]

Answer:

Solution A is a Weak Alkali, Solution B is a strong Acid.

Explanation:

At pH 10, the colour is blue, therefore it's a weak alkali.

At pH 1, the colour is red, therefore it's a strong Acid.

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A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i
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Answer:

- 8.89 m/s^{2}

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3 years ago
In a synthesis reaction, 2.43g of aluminum is reacted with 9.22g of chlorine. Show your work using a BCA table.
Zinaida [17]

Answer:

Chlorine gas is the limiting reactant

11.57 g

Explanation:

The equation of the reaction is;

2Al(s) + 3Cl2(g) -----> 2AlCl3(s)

The limiting reactant will produce the least number of moles of AlCl3.

For Al

Number of moles reacted = 2.43 g/27 g/mol = 0.09 moles

2 moles of Al yields 2 moles of AlCl3

0.09 moles of Al yields 0.09 * 2/2 = 0.09 moles of AlCl3

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9.22g/71g/mol = 0.13 moles

3 moles of Cl2 yields 2 moles of AlCl3

0.13 moles of Cl2 yields 0.13 * 2/3 = 0.087 moles of AlCl3

Hence Cl2 is the limiting reactant

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3 years ago
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