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erastova [34]
2 years ago
15

What is the empirical formula for a compound if a sample contains 1.0 g of S and 1.5 g of O?

Chemistry
2 answers:
RUDIKE [14]2 years ago
8 0
Moles S = 1/32
Moles O₂ = 1.5/32

simplest ratio = 1 : 1.5
= 2:3

Thus, empirical formula:
S₂O₃
The fourth option is correct.
Elenna [48]2 years ago
5 0
S atom is twice as heavy as an oxygen atom ( O is 16 and S is 32)
let's assume that 1 Arbitrary unit of sulfur is 1g heavy. this would mean that 2 units of Oxygen is 1g heavy(since it's half as heavy as S when comparing equal number of atoms). since we have 1.5g of oxygen, so there are 1.5*2=3 units of oxygen in that sample. and there are 1*1=1 units of Sulfur in that sample.
therefore the empirical formula is SO3
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Why carbon dioxide in the atmosphere appears to be increasing
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Explanation:

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2 years ago
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If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lu
kolbaska11 [484]

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1.505×10^23 atoms of lead

Explanation:

Volume of lead in the lungs = total volume of lungs = 5.60L

1 mole = 22.4L

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6 0
2 years ago
Water flowing at the rate of 13.85 kg/s is to be heated from 54.5 to 87.8°C in a heat exchanger by 54 to 430 kg/h of hot gas flo
weeeeeb [17]

Answer:

=> 572.83 K (299.83°C).

=> 95.86 m^2.

Explanation:

Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.

Hence;

Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.

15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).

If we solve for x, we will get the value of x to be;

x = 572.83 K (2.99.83°C).

x is the temperature of the exit gas that is 572.83 K(299.83°C).

(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).

∆T = 93.87/ln 1.38.

∆T = 291.521K.

Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.

heat-transfer area = 1931146.394/69.1 × 291.521.

heat-transfer area= 95.86 m^2.

7 0
2 years ago
Freon-11, CCl3F has been commonly used in air conditioners. It has a molar mass of 137.35 g/mol and its enthalpy of vaporization
PilotLPTM [1.2K]

Answer:

180.56 Kilo joules of energy is removed in the form of heat when 1.00 kg of freon-11 is evaporated.

Explanation:

Molar mass of freon-11 = 137.35 g/mol

Enthalpy of vaporization of freon-11= \Delta H_{vap}=24.8 kJ/mol

Mass of freon-11 evaporated = 1.00 kg = 1000 g

Moles of freon-11 evaporated = \frac{1000 g}{137.35 g/mol}=7.2807 mol

Energy in the form of heat removed when 1.00 kg of freon-11 gets evaporated:

7.28067 mol\times \Delta H_{vap}=7.2807 mol\times 24.8 kJ/mol

=180.56 kJ

0 0
2 years ago
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