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Nana76 [90]
3 years ago
10

Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G

iven that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have of Na present? a. 0.256 b. 256 c. 39.0 d. 0.0390 e. none of the above
Chemistry
1 answer:
seraphim [82]3 years ago
8 0

<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.

<u>Explanation:</u>

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g     (Conversion factor:  1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = \frac{100}{39}\times 0.1=0.256g of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

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7 0
3 years ago
38.0 mL of a 0.026 M solution of HCl is needed to react completely with a 0.032 M NaOH solution. Calculate the number of moles o
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<u>Answer:</u>

<em>The number of moles  of HCl actually present is 0.000988</em>

<u>Explanation:</u>

<em>The balanced chemical equation of the given reaction is </em>

HCl+NaOH \rightarrow NaCl+H_2 O

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we have to calculate the number of moles in 36 mL of HCl.

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5 0
4 years ago
Determine the number of milliliters of 0.00300 M phosphoric acid required to neutralize 40.00 mL of 0.00150 M calcium hydroxide.
a_sh-v [17]

The volume of H₃PO₄ : 13.33 ml

<h3>Further explanation</h3>

Given

0.003 M  Phosphoric acid-H₃PO₄

40 ml of 0.00150 M Calcium hydroxide-Ca(OH)₂

Required

Volume of H₃PO₄

Solution

Acid-base titration formula  

Ma. Va. na = Mb. Vb. nb  

Ma, Mb = acid base concentration  

Va, Vb = acid base volume  

na, nb = acid base valence  (amount of H⁺/OH⁻)

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Input the value :

a = H₃PO₄, b = Ca(OH)₂

0.003 x Va x 3 = 0.0015 x 40 x 2

Va = 13.33 ml

8 0
3 years ago
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