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OlgaM077 [116]
3 years ago
5

If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (2005 K)

Chemistry
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

205K

Explanation:

The following were obtained from the question:

n = 4moles

P = 5.6atm

V = 12L

R = 0.082atm.L/Kmol

T =?

PV = nRT

T = PV/nR

T = (5.6 x 12)/ (4 x 0.082)

T = 205K

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Solve 123,000 m x 3,234 m = ?
Leya [2.2K]

Answer:

397782000 m²

Explanation:

Given data:

123000 m × 3234 m =?

Solution:

123000 m × 3234 m = 397782000 m²

when we multiply the given quantities having same unit i.e meter the answer should be m² . It could be the area, and its unit is m².

Area = length (m) × width (m)

3 0
3 years ago
What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I
mihalych1998 [28]

<u>Answer:</u> The energy released in the given nuclear reaction is 94.99 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}

We are given:

Mass of _{92}^{235}\textrm{U} = 235.043924 u

Mass of _{0}^{1}\textrm{n} = 1.008665 u

Mass of _{53}^{131}\textrm{I} = 130.9061246 u

Mass of _{39}^{89}\textrm{Y} = 88.9058483 u u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(m_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.1019761u)\times c^2

E=(0.1019761u)\times (931.5MeV)  (Conversion factor:  1u=931.5MeV/c^2  )

E=94.99MeV

Hence, the energy released in the given nuclear reaction is 94.99 MeV.

3 0
3 years ago
What is possible reason for trial b not obeying the law of conservation of mass
IRISSAK [1]

The answer is D.)The balloon-like device was not used in this trial.

3 0
3 years ago
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How many moles of chlorine could be produced by decomposing 157g NaCl? 2NaCl --&gt; 2Na+Cl2
ss7ja [257]
Here is the answer for your question which the mole of chlorine is 1.34

5 0
2 years ago
A 7.27-gram sample of a compound is dissolved in 250 grams of benzene. The freezing point of this solution is 1.02°C below that
almond37 [142]

Answer:

146 g/mol → option b.

Explanation:

This is a problem about the freezing point depression. The formula for this colligative property is:

ΔT = Kf . m . i

We assume i = 1, so our compound is not electrolytic.

ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C

m  = molality (mol of solute/kg of solvent)

We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.

We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1

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0.19922 mol/kg = mol/ 0.250kg

mol = 0.19922 . 0.250kg → 0.0498 mol

molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol

5 0
3 years ago
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