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oksano4ka [1.4K]
3 years ago
14

A birthday balloon had a volume of 14.1 L when the gas inside was at a temperature of 13.9 °C. Assuming no gas escapes, what is

its volume when the balloon warms up to 22.0 °C?
Chemistry
1 answer:
bixtya [17]3 years ago
8 0

Answer:

14.5L

Explanation:

The following data were obtained from the question:

V1 = 14.1L

T1 = 13.9°C = 13.9 + 273 = 286.9K

T2 = 22°C = 22 + 273 = 295K

V2 =?

Using charles' law: V1/T1 = V2 /T2, we can obtain the new volume as follows:

14.1/286.9 = V2 /295

Cross multiply to express in linear form

286.9 x V2 = 14.1 x 295

Divide both side by 286.9

V2 = (14.1 x 295) / 286.9

V2 = 14.5L

Therefore, the new volume = 14.5L

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Answer:

°F - 32/1.8000

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What is amu of 99 % H-1, .2% H-1 and .8% H-3
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The average atomic mass of your mixture is 1.03 u .

The average atomic mass of H is the weighted average of the atomic masses of its isotopes.  

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).  

Thus,  

0.99    × 1.01 u = 0.998 u

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True or False. If a fire destroyed all of the grass in a grassland, this would NOT affect the biodiversity of this ecosystem.
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Answer:

This is false

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The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
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
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