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Rasek [7]
3 years ago
9

How well can you apply Charles’s law to this sample of gas that experiences changes in pressure and volume? Assume that pressure

and number of moles of gas are constant in this problem.
Chemistry
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

As temperature increases the volume of given amount of gas increases while pressure and number of moles remain constant.

Explanation:

According to the charle's law,

The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.

Mathematical expression:

V ∝ T

V = KT

V/T = K

When temperature changes from T₁ to T₂ and volume changes from V₁ to V₂.

V₁/T₁ = K        V₂/T₂ = K

or

V₁/T₁  = V₂/T₂

Thus, the ratio of volume and temperature remain constant for constant amount of gas at constant pressure.

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Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s
slava [35]

Answer:

Potassium carbonate (K₂CO₃)

Explanation:

The compounds dissociate into ions in water, as follows:

K₂CO₃ → 2 K⁺ + CO₃⁻    ⇒ 3 dissolved particles per mole

NaI → Na⁺ + I⁻    ⇒ 2 dissolved particles per mole

KBr → K⁺ + Br⁻   ⇒ 2 dissolved particles per mole

CH₃OH → CH₃O⁻ + H⁺  ⇒ 2 dissolved particles per mole

NH₄Cl → NH₄⁺ + Cl⁻   ⇒ 2 dissolved particles per mole

Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).

3 0
3 years ago
PLEASE HELP!!!
ladessa [460]

Answer:

the answer is c. [.4r]3d104324p

7 0
3 years ago
Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
3 years ago
Write an equilibrium equation that shows bisulfate acting as a weak acid in water.
NemiM [27]

Answer:

Explanation:

Bisulphate ion is a weak acid as it can form hydronium ion in water .

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

The equilibrium constant of this reaction is very small , hence bisulphate ion is very weak acid.

4 0
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Balance the chemical equation: <br> KOH + H3AsO4 → K2HAsO4 + H2O
Colt1911 [192]
2 <span>KOH +1 H3AsO4 →1 K2HAsO4 + 2 H2O</span>
3 0
3 years ago
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