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Leni [432]
2 years ago
14

How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction? (HCl molar mass is 36.46 g/mol).

Chemistry
1 answer:
Scilla [17]2 years ago
6 0

Answer:

Moles of BCl₃ needed = 0.089 mol

Explanation:

Given data:

Moles of BCl₃ needed = ?

Mass of HCl produced = 10.0 g

Solution:

Chemical equation:

BCl₃ + 3H₂O  →     3HCl + B(OH)₃

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 36.46 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of HCl with BCl₃.

               HCl             :           BCl₃

                 3               :             1

             0.27             :            1/3×0.27 = 0.089 mol

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A multivitamin tablet contains 40 milligrams of potassium. how many moles of potassium does each tablet contain?
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What is the hybridization of the central atom in the trichlorostannanide anion?
Mademuasel [1]

Answer:

sp²

Explanation:

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In the anion, the tin (Sn) forms 3 bonds, so, it has 1 electron that is isolated. According to the VSEPR theory, the geometry of a molecule depends on the bonded electrons and the lone pairs, because the repulsive force between them will form the angles of the molecule.

Thus, only the bonded electrons will contribute, and because of that, the geometry will be trigonal planar.

To form the bonds, the subshells (s, p, d, f) of the atoms in the bond must interact. But, only isolated electrons can bond, so, to be stable and form the compounds, some atoms have hybrids shells, which are formed by the join of subshells.

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8 0
3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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