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alexdok [17]
3 years ago
11

1. How much heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C?

Chemistry
1 answer:
diamong [38]3 years ago
8 0

Answer: An amount of 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

Explanation:

Given: mass of lead = 4.64 kg

Convert kg into grams as follows.

1 kg = 1000 g\\4.64 kg = 4.64 kg \times \frac{1000 g}{1 kg}\\= 4640 g

T_{1} = 150^{o}C

T_{2} = 219^{o}C

The standard value of specific heat of lead is 0.128 J/g^{o}C.

Formula used to calculate heat is as follows.

q = m \times C \times \Delta T

where,

q = heat energy

m = mass of substance

C = specific heat of substance

\Delta T = change in temperature

Substitute the value into above formula as follows.

q = m \times C \times \Delta T\\= 4640 g \times 0.128 J/g^{o}C \times (219 - 150)^{o}C\\= 40980.48 J/g^{o}C

Thus, we can conclude that 40980.48 J/g^{o}C heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

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1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t
slamgirl [31]

Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄  

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄  in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ?            ; c₂ = 2.0   mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500  mL of water.

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4 years ago
How many moles are contained in a 4.67-l sample of gas at 30 c and 199 kpa?
valentina_108 [34]

Using ideal gas equation,

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Answer:

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