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2 years ago

4.5 × 1025 atoms of nickel equal how many moles? A. 27 mol B. 2.7 × 1048 mol C. 0.75 mol D. 75 mol

Chemistry

2 answers:

crimeas [40]2 years ago

7 0

Answer:

75 mol

Explanation:

The number of atoms in a mole of substance is 6.023X10²³ [Avagadro's number]

if the given sample of nickel contains 4.5X10²⁵ atoms it means the moles are

moles = $\frac{numberofatoms}{Avagadro'snumber}$

moles = $\frac{4.5X10^{25}}{6.023X^{23}}=74.71mol$

If we round of the given answer the number of moles = 75 moles

Molodets [167]2 years ago

4 0

It's 75
PS use ^ to mean to the power of

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2 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

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Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

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<h3>10.57% of N in acetanilide</h3>

7 0

2 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

2 years ago

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Answer:

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3 0

3 years ago

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