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ArbitrLikvidat [17]
3 years ago
15

Which solution is the most concentrated? 2.0 mL of 10 M H2SO4, where H2SO4 has a molar mass of 98 g/mol 5.0 mL of 1.0 M PbSO4, w

here PbSO4 has a molar mass of 303 g/mol 2.0 mL of 10.5 M H2O2, where H2O2 has a molar mass of 34 g/mol 100 mL of 10 M NaCl, where NaCl has a molar mass of 58 g/mol
Chemistry
2 answers:
marishachu [46]3 years ago
7 0

Answer: the option 4) 2.0 mL of 10.5 M H₂O₂, where H₂O₂ has a molar mass of 34 g/mol.

Its concentration is 10.5 M.

Explanation:

1) The unit M means molar. It is the molarity of the solution.

Molartity is the concentration of the solution expressed as number of moles of solute per liters of solution.

The formula of molarity, M, is:

M = number of moles of solute / volume of solution in liters

2) 2.0 mL of 10 M H₂SO₄, where H₂SO₄ has a molar mass of 98 g/mol

⇒ concentration is 10 M

3) 5.0 mL of 1.0 M PbSO₄, where PbSO₄ has a molar mass of 303 g/mol

⇒ concentration = 1.0 M

4) 2.0 mL of 10.5 M H₂O₂, where H₂O₂ has a molar mass of 34 g/mol

⇒ concentration is 10.5 M

5) 100 mL of 10 M NaCl, where NaCl has a molar mass of 58 g/mol

⇒ concentration is 10 M

Mariana [72]3 years ago
3 0

Answer:

its c. 2.0 mL  of 10.5 M H2O2, where H2O2 has a molar mass of 34 g/mol

Explanation:

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Answer:  There is no question, but we can calculate a couple of items:

Density of sea water sample = (52.987g-44.317g)/8.5ml

Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample

Explanation:

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What is orbital theory of bonding in metals
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On the diagram above, trace the path of a light ray through these materials
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3 years ago
the concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock
julsineya [31]

Answer:

5.0 x 10⁹ years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of K-40 = 1.251 × 10⁹ years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (K-40) ([A₀] = 100%).

[A] is the remaining concentration of (K-40) ([A] = 6.25%).

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.

∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.

8 0
3 years ago
Read 2 more answers
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