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Lilit [14]
3 years ago
7

How much heat is required to raise the temperature of 250.0g of mercury 52.0°c if the specific heat of mercury is 0.140 j/(g x °

c)?
Chemistry
1 answer:
sasho [114]3 years ago
8 0

Q = m c T

c= 0.140 j/(g x °c)

m= 250.0g

T =52

hope you can solve it now

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There are several different models that represent compounds. One type of model is shown.
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Why are metals alloyed?
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5 0
2 years ago
Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
The density of lead is 11.4g/cm3. What volume, in ft3, would be occupied by 10.0 g of lead?
yawa3891 [41]

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

7 0
3 years ago
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