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ira [324]
3 years ago
10

Ionic compounds are solid at room temperature because they have

Chemistry
1 answer:
Bumek [7]3 years ago
5 0

Answer:option A High melting points

Explanation:

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If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume -  125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
3 years ago
If a sample of gas is located 10.2 cm from the injection point and the chart speed is 0.5 cm/min, what is the retention time?
Archy [21]

Answer : The retention time is, 20 min

Explanation :

Retention time : It is defined as the amount of time a compound spends on the column after it has been injected.

Formula of retention time is:

\text{Retention time}=\frac{\text{Distance from injection point to center of peaks}}{\text{Chart recorded speed}}

Given:

Distance from injection point to center of peaks = 10 cm

Chart recorded speed = 0.5 cm/min

Now put all the given values in the above formula, we get:

\text{Retention time}=\frac{10cm}{0.5cm/min}

Retention time = 20 min

Thus, the retention time is, 20 min

3 0
3 years ago
Which of the following is not a property of a buffer solution? a) It is a mixture of a weak acid and its conjugate base. b) Resi
m_a_m_a [10]

Answer:

e) pH is independent of concentration.

Explanation:

a) It is a mixture of a weak acid and its conjugate base. <em>TRUE. </em>A buffer is defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

b) Resists pH changes because it reacts with added acid or base. <em>TRUE. </em>Thermodynamically, the reaction of added acid or base is faster with the buffer mixture than with H⁺ or OH⁻ ions of the solutions.

c) The maximum buffer capacity is at pH = pKa. <em>TRUE. </em>The buffer capacity is pka±1. For this, buffer capacity is maximum in pka.

d) pH is dependent on the solution ionic strength and temperature. <em>TRUE.</em> Ionic strength and temperature are factors that influence concentrations of ions in solutions as the H⁺ ion that is the responsible

e) pH is independent of concentration. <em>FALSE. </em>pH in a buffer depends completely of concentrations of the acid and its conjugate base or vice versa.

I hope it helps!

8 0
3 years ago
Convert 78.9 kJ into calories (can you also explain how pls, ty &lt;3)
Katena32 [7]

Answer:

18857

Explanation:

1j=0.239cal

78.9kj=78900j

78.9kj= 78900×0.239=18857cal

4 0
3 years ago
50.0 g of copper was dropped into 250.0mL of H2O at 25.0°C in a coffee cup calorimeter. The copper and H2O reached thermal equil
siniylev [52]

Answer:

Initial temperature of copper = 163.3 °C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

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 = final temperature - initial temperature

Given data:

Mass of copper = 50 g

Initial temperature of copper = ?

Final temperature = 27.5°C

Initial temperature of water = 25.0°C

Specific heat capacity of copper = 0.385 j/g.°C

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 250.0 mL or 250.0 g

Solution:

Formula:

- Qm  =  +Qw

Now we will put the values in formula.

-50 g × 0.385 j/g.°C × [ 27.5°C -x] = 250 g × 4.184 J/g°C × [ 27.5°C - 25.0°C]

-50 g × 0.385 j/g.°C × [ 27.5°C -x] = 250 g × 4.184 J/g°C × 2.5°C

-19.25 j/°C × [ 27.5°C -x] = 2615 j

27.5°C -x =   2615 j / -19.25 j/°C

27.5°C -x = -135.8°C

x =  27.5°C + 135.8°C

x = 163.3 °C

7 0
3 years ago
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