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3 years ago

If 2.19 mol

Chemistry

1 answer:

never [62]3 years ago

3 0

Answer:

1370 °C

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data:

p = 8.12 atm

V = 36.41 L

n = 2.19

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

$\begin{array} {rcl}pV & = & nRT\\\text{8.12 atm} \times \text{36.41 L} & = & \rm\text{2.19 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\295.6&=&0.1797T\text{ K}^{-1}\\T& = &\dfrac{294.9 }{\text{0.1797 K}^{-1}}\\\\ & = & \text{1645 K}\\\end{array}$

T = (1645 - 273.15) °C = 1370 °C

The temperature of the sample is 1370 °C.

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Answer:

AS ACCORDING TO THE LAW OF MASS CONSERVATION

REACTANTS =PRODUCTS

THEREFORE,

223.4+96=MASS OF FE2O3

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3 years ago

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3 years ago

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Which substance is the limiting reactant when 4.0 g of sulfur reacts with 6.0 g of oxygen and 8.0 g of sodium hydroxide accordin

Marysya12 [62]

C, NaOH(aq) is the answer

6 0

3 years ago

A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

3 years ago

A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/

Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0

3 years ago