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3 years ago

The volume of carbon gas at 5.0 ATM was measured to be 363 ml. What will the pressure be if the volume is change to .00020 ml?

Chemistry

1 answer:

aleksley [76]3 years ago

5 0

Answer:

The answer to your question is P2 = 9075000 atm

Explanation:

Data

Pressure 1 = P1 = 5 atm

Volume 1 = V1 = 363 ml

Pressure 2 = P2 = ?

Volume 2 = 0.0002 ml

Process

To solve this problem use Boyle's law

P1V1 = P2V2

-Solve for P2

P2 = P1V1/V2

-Substitution

P2 = (5 x 363) / 0.0002

-Simplification

P2 = 1815 / 0.0002

-Result

P2 = 9075000 atm

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Answer:

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It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principall

Aleks [24]

Answer:

Hydrogen: -141 kJ/g

Methane: -55kJ/g

The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.

Explanation:

According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qc + Qb = 0

Qc = -Qb [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in the temperature

<h3>Hydrogen</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ

The heat released per gram of hydrogen is:

$\frac{-162kJ}{1.15g} =-141 kJ/g$

<h3>Methane</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ

The heat released per gram of methane is:

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3 years ago

Assuming that both cases describe hydrogen‑like atoms with one electron, for which case is more energy emitted or absorbed?

Nana76 [90]

Assuming that both cases describe hydrogen‑like atoms with one electron, More energy is emitted or absorbed for case 2. The correct option is D.

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The energy of the electron decreases as it changes levels, and emission of photons happens in the atom.

With the electron moving from a higher to a lower energy level, the photon is emitted. The photon's energy is the same as the energy lost by an electron moving to a lower energy level.

Thus, the correct option is D, More energy is emitted or absorbed for case 2.

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1 year ago

Peut-on être exact sans être précis?

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2 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

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