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2 years ago

2. What is the temperature (°C) of 1.50 moles of gas stored in a 10.0 L container at 1558 mm

Chemistry

1 answer:

nalin [4]2 years ago

6 0

Answer:

-106°C

Explanation:

1558 mmHg / 1 × 1 atm / 760. mmHg = 2.05 atm

PV = nRT

2.05 atm (10.0 L) = 1.50 mol (0.082 atmL/molK) T

20.5 atmL = (0.123 atmL/K) T

K/0.123 atm L × 20.5 atmL = (0.123 atmL/K) T × K/0.123 atmL

166.666 K = T

167 K = T (sig figs)

K = °C + 273

167 - 273 = °C - 273

-106 = °C

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Answer:

The answer is: (a) positive; (b) negative.

Explanation:

The change in enthalpy (ΔH) of a reaction is the amount of energy absorbed or released during a chemical reaction carried out at constant pressure.

a) In an endothermic chemical reaction, heat energy is absorbed by the system from the surrounding. Therefore, the sign of enthalpy change for an endothermic process is positive, ΔH= positive.

b) In an exothermic chemical reaction, heat energy is released by the system into the surrounding. Therefore, the sign of enthalpy change for an exothermic process is negative, ΔH= negative.

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3 years ago

A quantity of water is heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy. If the specific heat of water is 4.18 J /

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Answer:

6,8 g

Explanation:

c = 4.18 J/(g * °C) = 4180 J / (kg * °C)

$t_{1}$ = 25 °C

$t_{2}$ = 36,4 °C

Q = 325 J

The formula is: Q = c * m * ( $t_{2} - t_{1}$ )

m = $\frac{Q}{c * (t_{2} - t_{1} )}$

Calculating:

m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g

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3 years ago

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c) Atomic Mass.

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3 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

2 years ago

Read 2 more answers