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3 years ago

What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?

1 answer:

5 0

Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = 0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

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An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron? A. 450.00

valkas [14]

Answer:

Option A. 450.00

Explanation:

1) Data:

a) m = 18 kg

b) T₁ = 285 K

c) T₂ = 318 K

d) Q = 267.3 kJ

e) S = ?

2) Principles and equations

The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.

The mathematical relation between the specific heat and the heat energy absorbed is:

Q = m × S × ΔT

Where,

Q is the heat absorbed,
S is the specific heat, and
ΔT is the temperature increase (T₂ - T₁)

3) Solution:

a) Substitute the data into the equation:

267.3 kJ = 18 kg × S × (318 K - 285 K)

b) Solve for S and compute:

S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)

The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.

c) Convert to J /( kg . k):

0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)

Now we can see that the option A is is the answer, assuming the units.

6 0

3 years ago

while doing an experiment you got stuck and didn't know how to proceed what will you do to continue with the experiment

stiv31 [10]

Answer:

Look for extra things to do, small details, until you find a big enough one to go off that to continue

Explanation:

n/a

7 0

3 years ago

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

What would the volume of a gas be at 150c if had of volume of 693 ml at 45 c

Vlad1618 [11]

Answer:

Explanation:

T1 = 150°C = (150 + 273.15)K = 423.15K

T2 = 45°C = (45 + 273.15)K = 318K

V1 = 693mL = 693cm³

Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.

V = kT

V1 / T1 = V2 / T2

693 / 423.15 = V2 / 318

V2 = (693 * 318) / 423.15 = 520.79cm³

The new volume of the gas is 520.79cm³

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3 years ago

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erik [133]

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3 years ago