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3 years ago

At 1.4 atm of pressure a gas has a volume of 111 ML at what pressure will the gas have a volume of 85 ML.

Chemistry

1 answer:

Alona [7]3 years ago

8 0

Answer:

The pressure will be 1, 83 atm

Explanation:

We use Boyle Mariotte's formula: for a given mass of gas at constant temperature, the pressure and volume vary inversely proportionally. We convert the unit ml into L: 111/1000= 0,111 L and 85/1000= 0,085 L.

P1xV1=P2xV2

P2= (P1xV1)/V2=(1,4 atm x 0,111L)/0,085L=<em>1,83 atm</em>

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3 years ago

Compared to ultraviolet light, an electromagnetic wave that has a higher frequency will also have ________.

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3 years ago

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3 years ago

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Merits of modern periodic table?

ANEK [815]

Answer:

<h3>Merits of modern periodic table:</h3>

The wrong position of some elements like argon, potassium, cobalt and nickel due to atomic weights have been solved by arranging the elements in the order of increasing atomic number without changing their own places.
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6 0

3 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

Lady bird [3.3K]

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:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

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

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

<em>Moles H₂SO₄ added:</em>

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

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

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

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

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

7 0

2 years ago