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3 years ago

Why is it necessary to specify temperature when giving Ka value

Chemistry

1 answer:

Yuliya22 [10]3 years ago

5 0

Answer:

Because the equilibrium constant is only at constant at temperature. Changing the temperature changes the Ka.

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The element californium (cf) sells for $1000 per µg. assuming 6.02 x 1023 atoms of cf have a mass of 251 grams, how many atoms o

saveliy_v [14]

Answer:- $2.40*10^1^0atoms$

Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.

We know that, 1 US dollar = 100 cents

1 cent = 1 US penny

So, 1 US dollar = 100 US pennies

$1g=10^6\mu g$

Let's make the set up starting with 1 penny as:

$1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})$

= $2.40*10^1^0atoms$

Therefore, we can bye $2.40*10^1^0atoms$ of Cf in one US penny.

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3 years ago

Which accurately completes the statement? Conduction involves the transfer of ___________ due to the movement of particles. Grou

Jlenok [28]

Explanation:

conduction involves the transfer of molecules due to the movement of particles

5 0

3 years ago

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

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3 years ago

4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate

Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

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3 years ago

The temperature of a sample of Rn is changed without changing its volume, causing a change in pressure from 8.6731 atm to 0.5761

Flauer [41]

Answer:

yeye

Explanation:

3 0

3 years ago

Why is it necessary to specify temperature when giving Ka value​

Why is it necessary to specify temperature when giving Ka value