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Lelechka [254]
3 years ago
9

Calculate the amount of heat required to completely sublime 55.0 g of solid dry ice CO2 at its sublimation temperature. The heat

of sublimation for carbon dioxide is 32.3 kj mol
Chemistry
1 answer:
aliina [53]3 years ago
7 0

Answer:

40.4 kJ

Explanation:

Step 1: Given data

  • Mass of CO₂ (m): 55.0 g
  • Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol

Step 2: Calculate the moles corresponding to 55.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

n = 55.0 g × 1 mol/44.01 g = 1.25 mol

Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂

We will use the following expression.

Q = n × ΔH°sub

Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ

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All the noble gases have similar chemical properties. <br> True or false?
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3 years ago
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
cricket20 [7]

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10  

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10  + O2---------------->  CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF  CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE  IT IS A

8:2 OR 4:1 RATIO.  THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

SO

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

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6 0
3 years ago
A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h
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Answer:

5.758  is the density of the metal ingot in grams per cubic centimeter.

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Density of water at 20 °C = d =998.2 kg/m^3

1 kg = 1000 g

1 m^3=10^6 cm^3

998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3

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Volume of pycnometer = Volume of water present in it = V

Density=\frac{Mass}{Volume}

V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' =  9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3

Density of metal = d'

Volume of metal = v' =\frac{M'}{d'}

Difference in volume will give volume of metal ingot.

v' = v - V

v'=19.73 cm^3-18.08 cm^3=

v'=1.65 cm^3

Since volume cannot be in negative .

Density of the metal =d'

=d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3

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Answer:

C I'm pretty sure

Explanation:

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