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2 years ago

2.0 L of Carbon dioxide is heated from -25.0 °C to Standard Temperature.

Chemistry

1 answer:

RideAnS [48]2 years ago

5 0

The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume
V2 = final volume
T1 = initial temperature
T2 = final temperature

V1 = 2L
V2 = ?
T1 = -25°C + 273 = 248K
T2 = 273K

2/248 = V2/273

273 × 2 = 248V2

546 = 248V2

V2 = 546/248

V2 = 2.2L

Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L

Learn more about volume at: brainly.com/question/11464844

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The gram formula mass of a compound is 48 grams. The mass of 1.0 moles of this compound is

ehidna [41]

The gram formula mass is Molar mass. The mass of 1.0 moles is :
3) 48.0 g

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3 years ago

Carbon-14 has a half-life of 5,730 years. In how many years will 120 grams of Carbon-14 decay to 15 grams? 11,460 years 17,190 y

erastova [34]

120 grams of Carbon-14 decays to 15 grams in 17,190 years.

7 0

3 years ago

A sample of table sugar (sucrose c12h22o11) has a mass of 3.115 g

attashe74 [19]

Hey there:

a) atomic mass:

Carbon =12.0107 g/mol

Hydrogen = 1.00794 g/mol

Oxygen = 15.9994 g/mol

Therefore:

C12H22O11 =

12 * 12.0107 + 1 * 1.00794 + 16 * 15.9994 => 342.29648 g/mol

__________________________________________________________

b) number of moles:

n = m / mm

n = 3.115 / 342.29648

n = 0.0091 moles

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hope this helps!

6 0

3 years ago

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H

Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

where,

$\Delta T_f$ = change in freezing point

$k_f$ = freezing point constant (for benzene} = $5.12^0Ckg/mol$

m = molality

Putting in the values we get:

$0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}$

${\text{ Mass of solute in g}}=0.028g$

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

4 0

3 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

2 years ago