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Kaylis [27]
3 years ago
8

How many atoms of the element with atomic number 11 are in the reactants of the given formula?

Chemistry
1 answer:
liq [111]3 years ago
3 0

The number of Na atoms in the reactants = 2

<h3>Further explanation</h3>

Every chemical reaction involves compounds that can act as reactants or products

The reactants will react to produce products

The reactants are usually written on the left side of the reaction, while the products on the right side of the reaction

Reaction

2 NaCl + H₂SO₄ → 2 HCl + Na₂SO₄

The element with atomic number 11  = Na(Natrium)

The number of atoms of the element that reacts or is produced can be seen from the reaction coefficient and the number of atoms in the compound

Na in NaCl = 1 atom

From equation, cofficient for NaCl = 2, so number of atoms of Na=2 x 1 = 2

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An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
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Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

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M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

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