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3 years ago

What mass, in grams, of sodium bicarbonate, nahco3, is required to neutralize 1000.0 l of 0.350 m h2so4?

1 answer:

3 0

Hello!

The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:

2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂

To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:

$1000 L H_2SO_4* \frac{0,350 moles H_2SO_4}{1 L}* \frac{2 moles NaHCO_3}{1 mol H_2SO_4}* \frac{84,007gNaHCO_3}{1molNaHCO_3} \\ \\ =58804,9 g NaHCO_3$

So, to neutralize 1000 L of 0,350 moles of H₂SO₄ you'll need 58804,9 grams of NaHCO₃

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When you have finished adding the necessary amount of NaOH to your beaker, what color will the litmus paper turn?

Svetllana [295]

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Explanation:

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7 0

3 years ago

What is the molar concentration of

Leto [7]

Answer:

1.42 M

Explanation:

First calculate the amount of moles.

that's done by dividing the mass with the molecular mass so 660g / 310.18 g/mol = 2.13 mol

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(without rounding: 1.418531175 M)

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2 years ago

Hydrogen 3 has a half life of 12.32 years a sample of h-3 weighing 3.02 grams is left for 15.0 years what will the final weight

yawa3891 [41]

Answer:

The final mass of sample is 1.3 g.

Explanation:

Given data:

Half life of H-3 = 12.32 years

Amount left for 15.0 years = 3.02 g

Final amount = ?

Solution:

First all we will calculate the decay constant.

t₁/₂ = ln² /k

t₁/₂ =12.32 years

12.32 y = ln² /k

k = ln²/12.32 y

k = 0.05626 y⁻¹

Now we will find the original amount:

ln (A°/A) = Kt

ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y

ln (3.02 g/ A) = 0.8439

3.02 g/ A = e⁰°⁸⁴³⁹

3.02 g/ A = 2.33

A = 3.02 g/ 2.33

A = 1.3 g

The final mass of sample is 1.3 g.

8 0

3 years ago

Read 2 more answers