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Gelneren [198K]
3 years ago
6

To make 2.00 L of 0.300 M sulfuric acid, how many mL of a 1.00 M stock solution should be used?

Chemistry
1 answer:
Tamiku [17]3 years ago
3 0
We are given with
M1 = 1.00 M

M2 = 0.300 M
V2 = 2.00 L

We are asked to get V1

Using material balance
M1 V1 = M2 V2
Substituting the given values
1.00 V1 = 0.300 M (2.00 L)
V1 = 0.600 L or 600 mL
THe volume needed is 600 mL<span />
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bija089 [108]

Answer:

D) HCl(aq)

Explanation:

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In this scenario, the chemical formula which represents a homogeneous mixture is aqueous hydrogen chloride, HCl(aq). The aqueous hydrogen chloride is a homogeneous mixture of water and hydrogen chloride. This ultimately implies that, aqueous hydrogen chloride HCl(aq) is a solution of hydrogen chloride in water and it is commonly referred to as Hydrochloric acid.

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HCl(aq) + H_{2}O(l)  →  H_{3}O^{+} (aq) + Cl_{−}(aq)

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Sonbull [250]

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Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.
lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

<u>Step 1</u>: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

<u>Step 2:</u> The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

<u>Step 3: </u>Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) +  I-(aq)

<u>Step 4</u>: Calculate iodide ions

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<u>Step 5:</u> Calculate molarity of iodide ion

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Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0
3 years ago
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4 0
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