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Hatshy [7]
3 years ago
13

Balance the following equation mg(s)+ HCl H2)g)+MgCl2

Chemistry
1 answer:
Marat540 [252]3 years ago
6 0
<h3>Given equation:-</h3>

\\ \sf\longmapsto Mg+HCl\longrightarrow MgCl_2+H_2

<h3>BALANCED EQUATION:-</h3>

\\ \sf\longmapsto Mg+2HCl\longrightarrow MgCl_2+H_2

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If 0.100 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams o
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Answer:

15.6g Ag2SO4

Explanation:

2AgNO3 + H2SO4 --> Ag2SO4 + 2HNO3

-2x              -x

0.1-2x.       0.155-x

x=0.05      x=0.155

0.05mol Ag2SO4 x 311.78g = 15.6g Ag2SO4

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3 years ago
How many significant digits are in the measurement 64,000 mg<br> A.2 <br> B.3<br> C.4<br> D.5
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What happens when the temperature of a solid reaches it’s melting point
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R-CH2-COOH in presence of X2 / red P and H2O gives ________<br>Name the reaction​
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8 0
2 years ago
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate
AlladinOne [14]

The given question is incomplete. The complete question is as follows.

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When Ba^{2+} and Na_{2}SO_{4} are added  then white precipitate forms. And, reaction equation for this is as follows.

       Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of BaSO_{4} is 233.43.

Now, we will calculate the number of moles as follows.

  No. of moles = mass × M.W

                        = \frac{0.212}{233.43}

                        = 0.00091 mol of BaSO_{4}

Hence, it means that 0.00091 mol of Ba^{2+}. Now, we will calculate the mass as follows.

       Mass = moles × MW

                 = 0.00091 \times 137.327

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Thus, we can conclude that mass of barium into the original solution is 124 mg.

8 0
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