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Tems11 [23]
2 years ago
7

Arrange the following compounds from left to right in order of increasing percentage by mass of hydrogen: a) H2O, b) C12H26, c)

N4H6, d) LiH.
Chemistry
1 answer:
Nuetrik [128]2 years ago
4 0

Answer:

(A) N4H6  (B) H2O  (C) LiH  (D) C12H26

Explanation:

The given compounds have been arranged  from left to right in order of increasing percentage by mass of hydrogen.

The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.

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Answer:

<u>Formula</u><u>:</u> Velocity \:  V = f \lambda \\  Solution: = 5 \times 0.8 \\  = 4 \:  {ms}^{ - 1}

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If a sample of N2 gas has an initial pressure of 500 Torr and volume of 0.5 L, what will the final volume be if the pressure is
kherson [118]

Answer:

V₂ = 0.4 L

Explanation:

<u>Data:</u>

  • P₁ (initial pressure) = 500 torr
  • V₁ (initial volume) = 0.5 L
  • P₂ (final pressure) = 700 torr

<u>Wanted:</u>

  • V₂ (final volume)

<u>Equation:</u>

  • P₁V₁ = P₂V₂  →  V₂ = P₁V₁ / P₂

<u>Solution:</u>

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2 years ago
When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

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