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vagabundo [1.1K]
3 years ago
5

A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution before the addition of any lioh

.
Chemistry
1 answer:
snow_lady [41]3 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

 <span>moles HClO4 = 0.100 L x 0.18 M = 0.018 
moles LiOH = 0.030 L x 0.27 = 0.0081 
moles H+ in excess = 0.018 - 0.0081 = 0.0099 
total volume = 0.130 L 
[H+] = 0.0099/ 0.130= 0.0762 M 
pH = 1.12</span>
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how much heat, in terms in q, would it take to produce the change in temperature indicated in the picture? what is your reasonin
STALIN [3.7K]

Answer:

1. q.

2. 2q.

3. 3q.

4. 6q.

Explanation:

We'll begin by calculating the specific heat capacity of the liquid. This can be obtained as follow:

Mass (m) = 25 g

Change in temperature (ΔT) = 20 °C

Heat (Q) = q

Specific heat capacity (C) =?

Q = MCΔT

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C = q/500

C = 2×10¯³ qg°C

Therefore, the specific heat capacity of liquid is 2×10¯³ qg°C

Now, we shall determine the heat required to produce the various change in temperature as follow:

2. Mass (m) = 50 g

Change in temperature (ΔT) = 20 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 20

Q = 2q.

Therefore, the heat required is 2q.

3. Mass (m) = 25 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 25 × 2×10¯³ × 60

Q = 3q.

Therefore, the heat required is 3q.

4. Mass (m) = 50 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 60

Q = 6q.

Therefore, the heat required is 6q.

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3 years ago
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ruslelena [56]

Answer:

A

Explanation:

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4 0
3 years ago
How many photons are produced in a laser pulse of 0.364 J at 477 nm?
Luda [366]

Answer:

1.00 × 10¹⁸

Explanation:

1. Calculate the <em>energy of one photon</em>

The formula for the energy of a photon is

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λ = 477 nm = 477 × 10⁻⁹ m                              Insert the values

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2. Calculate the <em>number of photons</em>

Divide the total energy by the energy of one photon.

No. of photons = 0.418 × 1/4.165 × 10⁻¹⁹      

No. of photons = 1.00 × 10¹⁸

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34kurt
This is known as polymerisation 
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