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Snowcat [4.5K]
3 years ago
14

A)How many moles of O2 are required for the complete combustion of 2.2 g of C3H8 to form CO2 and H2O?

Chemistry
1 answer:
Oksana_A [137]3 years ago
4 0
<span>A)How many moles of O2 are required for the complete combustion of 2.2 g of C3H8 to form CO2 and H2O?

C3H8 + 5O2 = 3CO2 + 4H2O

2.2 g C3H8 ( 1 mol / 44.11 mol ) ( 5 mol O2 / 1 mol C3H8) = 0.25 mol O2


b)A 65.25 g sample of CuSO4•5H2O (M = 249.7) is dissolved in enough water to make 0.800 L of solution. What volume of this solution must be diluted with water to make 1.00 L of 0.100 M CuSO4?

M1V1 = M2V2
[65.25 ( 1 / 249.7 )/ 0.800 L] (V1)= 0.100  (1.00 L)
V1 = 0.31 L or 310 mL </span>
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Radioisotopes are unstable because they contain an excess number of:.
myrzilka [38]
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A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an
MrRa [10]

Answer : 0.026 moles of oxygen are in the lung

Explanation :

We can solve the given question using ideal gas law.

The equation is given below.

PV = nRT

We have been given P = 21.1 kPa

Let us convert pressure from kPa to atm unit.

The conversion factor used here is 1 atm = 101.3 kPa.

21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm

V = 3.0 L

T = 295 K

R = 0.0821 L-atm/mol K

Let us rearrange the equation to solve for n.

n = \frac{PV}{RT}

n = \frac{0.208atm\times 3.0L}{0.0821 L.atm/mol K\times 295 K}

n = 0.026 mol

0.026 moles of oxygen are in the lung

3 0
3 years ago
A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the
Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

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Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q =  m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C

-4784 * c(aluminium) = -4184

c(aluminium) = 0.875 J /g°C

The specific heat of aluminium is 0.875 J/g°C

7 0
3 years ago
Which branch of chemistry is most likely to involve the study of the earths crust
stepan [7]
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Roland is measuring a physical property of an object. He records the measurement:
Vsevolod [243]

Answer:

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Explanation:

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