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nordsb [41]
2 years ago
8

If 75 grams of oxygen react, how many grams of aluminum are required?

Chemistry
1 answer:
german2 years ago
5 0

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

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Which of the following polyatomic ions will form an ionic compound with two sodium ions? CO32− HCO31− NO21− NO31−
BaLLatris [955]

Answer:

CO32−

Explanation:

We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.

When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;

2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)

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3 years ago
19. What is sthe mass of 6.02 x 1024 atoms
alekssr [168]

The mass of magnesium in 6.02 \times 10^{24} atoms is 240 g.

Answer: Option A

<u>Explanation:</u>

First, we have to convert the atoms to moles of magnesium.

We know that 6.02 \times 10^{24} atoms are present in 1 mole of magnesium. So,

    \text { 1 atom }=\frac{1}{6.02 \times 10^{24}} \text { moles }

    6.02 \times 10^{24} \text { atoms }=\frac{6.02 \times 10^{24}}{6.02 \times 10^{23}} \text { moles }=10 \text { moles }

Thus,

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