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kati45 [8]
2 years ago
11

If 100mL was diluted of a 2.0 M solution of sodium chloride by adding an additional 100 mL of water, what would the new molarity

be?
Chemistry
1 answer:
andrew-mc [135]2 years ago
7 0
I need help with my questions sorry
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Consider the balanced equation. 2HCl + Mg MgCl2 + H2 If 40.0 g of HCl react with an excess of magnesium metal, what is the theor
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Answer:

Theoretical yield of hydrogen is 1.11 g

Explanation:

Balanced equation, Mg+2HCl\rightarrow MgCl_{2}+H_{2}

As Mg remain present in excess therefore HCl is the limiting reagent.

According to balanced equation, 2 moles of HCl produce 1 mol of H_{2}.

Molar mass of HCl = 36.46 g/mol

So, 40.0 g of HCl = \frac{40.0}{36.46}moles of HCl = 1.10 moles of HCl

Hence, theoretically, number of moles of H_{2} are produced from 1.10 moles of HCl = (\frac{1}{2}\times 1.10)moles=0.550moles

Molar mass of H_{2} = 2.016 g/mol

So, theoretical yield of H_{2} = (0.550\times 2.016)g=1.11g

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Identify the balanced equation for the following reaction:<br><br> SO2(g) + O2(g) → SO3(g)
sergiy2304 [10]

Answer:  The balanced equation for the given reaction is

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)

Here, number of atoms on reactant side are as follows.

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Number of atoms on product side are as follows.

  • S = 1
  • O = 3

To balance this equation, multiply SO_{2} by 2 on reactant side and multiply SO_{3} by 2. Hence, the equation will be re-written as follows.

2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)

Here, number of atoms on reactant side are as follows.

  • S = 2
  • O = 6

Number of atoms on product side are as follows.

  • S = 2
  • O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is 2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g).

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¿es demasiado tarde para ayudarte?

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