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3 years ago

What volume of 3.00 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M Na2CO3?

Chemistry

1 answer:

Alla [95]3 years ago

5 0

Answer:

0.0500L of 3.00M HCl are needed to react completely

Explanation:

Hydrochloric acid HCl reacts with sodium carbonate, Na2CO3, as follows:

2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂

<em>Where 2 moles of HCl reacts per mole of Na₂CO₃</em>

<em />

To find the volume of 3.00M HCl needed to react completely we need first to calculate the moles of Na₂CO₃ added and, with this chemical equation we can solve for moles of HCl as follows:

<em>Moles Na₂CO₃:</em>

0.750L * (0.100moles Na₂CO₃ / L) = 0.0750moles

<em>Moles HCl:</em>

0.0750moles Na₂CO₃ * (2 moles HCl / 1 mole of Na₂CO₃) = 0.150 moles of HCl

<em>Volume of 3.00M HCl:</em>

0.150 moles of HCl * (1L / 3.00moles) =

<h3>0.0500L of 3.00M HCl are needed to react completely</h3>

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A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o

azamat

The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg

5 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

4 years ago

Is sugar (C12H22O11) an element or a compound?

Semmy [17]

I believe that sugar is a compound because there are elements that make up sugar

4 0

4 years ago

A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 293 K at a pressure of

Mekhanik [1.2K]

Answer:

2.42L

Explanation:

Given parameters:

V₁ = 1.8L

T₁ = 293K

P₁ = 101.3kPa

P₂ = 67.6kPa

T₂ = 263K

Unknown:

V₂ = ?

Solution:

To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

All the units are in the appropriate form. We just substitute and solve for the unknown;

101.3 x 1.8 / 293 = 67.6 x V₂ / 263

V₂ = 2.42L

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3 years ago

What type of bacteria convert the

Ksju [112]

Answer: B

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3 years ago