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Nostrana [21]
2 years ago
6

. How many grams of Na2SO4 are required to make 2700 mL of a 2.0 M solution?

Chemistry
2 answers:
ratelena [41]2 years ago
6 0

Answer:

Explanation:

From the net ionic equation

Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4

Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+

Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-

Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)

Andru [333]2 years ago
6 0

Answer: Approx 2kg of salt are required.

Explanation: If 4.5 ⋅ L of a 3.0 ⋅ m o l ⋅ L − 1 solution are required, this is a molar quantity of 4.5 ⋅ L × 3.0 ⋅ m o l ⋅ L − 1 = 13.5 ⋅ m o l . Now, we simply multiply this molar quantity by a molar mass to get the mass: 13.5 ⋅ m o l × 142.04 ⋅ g ⋅ m o l − 1 = ? ? ⋅ g

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Given the reaction 2NaOH + H2SO4 = Na2SO4 + 2H2O, what is the total number of grams of NaOH needed to react completely with 196
ollegr [7]

Answer:

160 g of NaOH

Explanation:

Reaction: 2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

When a base and an acid react, the produce water and a salt. → Neutralization

In this reaction 2 moles of hydroxide need 1 mol of sulfuric acid to react, it is the same to say that 1 mol of sulfuric acid needs 2 moles of NaOH to react. We convert the mass of sulfuric to moles and we propose the rule of three:

196 g / 98 g/mol = 2 moles

1 mol of sulfuric acid needs 2 moles of NaOH to react

Then 2 moles of sulfuric acid will react with (2 . 2) /1 = 4 moles of NaOH

We convert the moles of base to grams → 4 mol . 40 g /1mol = 160 g

3 0
3 years ago
Which property describes the shininess of an element?
goldenfox [79]

Answer: Metals are shiny and lustrous with a high density. They have very high melting and boiling points because metallic bonding is very strong, so the atoms are reluctant to break apart into a liquid or gas.

Explanation:

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6 0
3 years ago
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Which one is the least dangerous to humans?
3241004551 [841]

Answer:

A is correct

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Explanation:

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7 0
2 years ago
PLEASE HELP!!! PLEASE.
adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄:  The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

<u><em>Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl? </em></u>

  • As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
  • The normality of the NaOH and HCl = Their molarity.

<em>∵ (NV)NaOH = (NV)HCl</em>

<em>∴ N of HCl = (NV)NaOH / (V)HC</em>l = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

<u><em>Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?</em></u>

  • As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
  • <em>The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.</em>

<em>∵  (NV)H₂SO₄ = (NV)KOH</em>

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

<em>∴ The concentration of KOH = 7.675 mN = 7.675 mM.</em>

<em></em>

<u><em>Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?</em></u>

  • As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

<em>The normality of NaOH = Molarity of NaOH = 0.5 N.</em>

<em>∵ (NV)H₂SO₄ = (NV)NaOH</em>

<em>∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄</em> = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

<em>∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.</em>

<em></em>

<u><em>Q₄: Explain the difference between an endpoint and equivalence point in a titration.</em></u>

  • The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
  • The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
  • End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
  • The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
6 0
3 years ago
What element is 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f10
skad [1K]
Answer:

Bi (Bismuth)
4 0
3 years ago
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