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4 years ago

Calculate the number of kilojoules to warm 125 g of iron from 23.5°C to 78.0°C.

1 answer:

6 0

M = 125 g = 125/1000 kg = 0.125 kg, θ₂ = 78.0°C, θ₁ = 23.5°C , c = 0.44kJ/kg°C,

Q = mc(θ₂ - θ₁)

Q = 0.125*0.440*(78 - 23.5)

Q = 2.9975 kJ

2.9975 kJ<span> required to warm it.</span>

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g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u

DIA [1.3K]

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg . 0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

4 0

4 years ago

Glycerol (molar mass 92.09 g/mol) has been suggested for use as an alternative fuel. The enthalpy of combustion, Δ

victus00 [196]

The mass of glycerol to that would need to be combusted to heat 500.0g of water from 20.0°C to 100.0°C is; 9.32 grams.

We must establish the fact that energy is neither created nor destroyed.

Therefore, the amount of heat absorbed by water is equal to the amount of heat released by the combustion of glycerol.

Total heat absorbed by water, H(water) is;

Q(water) = m C (T2 - T1)

Q(water) = 500 × 4.184 × (100-20)

Q(water) = 167.36 kJ

Consequently, the quantity of heat evolved by the combustion of glycerol is;

Q(glycerol) = 167,360 J = n × ΔH°comb

where, n = no. of moles of glycerol.

167.36 kJ= n × 1654 kJ/mole

n = 167.36/1654

n = 0.1012 moles of glycerol.

Therefore, mass of glycerol combusted, m is;

m = n × Molar mass

m = 0.1012 × 92.09

m = 9.32 g.

brainly.com/question/20709115

6 0

3 years ago

List earth and the gas giant planets from the hottest to the coldest planet

inessss [21]

Answer:

1. Venus

471°C

2. Mercury

(430°C) during the day, (-180°C) at night

3. Earth

16°C

4. Mars

-28°C

5. Jupiter

-108°C

6. Saturn

-138°C

7. Uranus

-195°C

8. Neptune

-201°C

Explanation:

8 0

2 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

The formation of a white cloud

saw5 [17]

Most clouds are white. That's because water and ice particles that make up a cloud have just the right amount and sizes to scatter light in all possible wavelengths. When light of practically all wavelengths combine, the result is white light.

4 0

3 years ago