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3 years ago

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A balloon contains helium gas expands from 230ml to 860 ml as more helium is added. What was the initial quantity of helium pres

ent if the expanded balloon contains 3.8x10^-4 mol, assuming constant temperature and pressure?

2 answers:

Kisachek [45]3 years ago

6 0

Answer:

n₁ = 1.0× 10⁻⁴ mol

Explanation:

Given data:

Initial volume of balloon = 230 mL

Initial number of moles of He =?

Final number of moles of He = 3.8 × 10⁻⁴ mol

Final volume of balloon = 860 mL

Solution:

The given problem will be solve through Avogadro law,

"Number of moles of gas and volume are directly proportional to each other at constant temperature and constant pressure"

Mathematical relationship:

V₁/n₁ = V₂/n₂

No we will put the values.

230 mL /n₁ = 860 mL/ 3.8 × 10⁻⁴ mol

n₁ = 230 mL× 3.8 × 10⁻⁴ mol/ 860 mL

n₁ = 874 × 10⁻⁴ mol. mL / 860 mL

n₁ = 1.0× 10⁻⁴ mol

PolarNik [594]3 years ago

3 0

Answer: the answer is n1=n2v1/v2

Explanation:

hope this helped

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An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine

nekit [7.7K]

Answer:

The molecular formula of the compound is $C_{7}H_{6}O_{2}$ .

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

Carbon

$m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{C} = 84.067\,g$

Hydrogen

$m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{H} = 6.057\,g$

Oxygen

$m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{O} = 31.995\,g$

Now, the number of moles ( $n$ ), measured in moles, of each element are calculated by the following expression:

$n = \frac{m}{M}$

Where:

$m$ - Mass of the element, measured in grams.

$M$ - Molar mass of the element, measured in grams per mol.

Carbon ( $m_{C} = 84.067\,g$ , $M_{C} = 12.011\,\frac{g}{mol}$ )

$n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }$

$n = 7$

Hydrogen ( $m_{H} = 6.057\,g$ , $M_{H} = 1.008\,\frac{g}{mol}$ )

$n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }$

$n = 6$

Oxygen ( $m_{O} = 31.995\,g$ , $M_{O} = 15.999\,\frac{g}{mol}$ )

$n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }$

$n = 2$

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

$C_{7}H_{6}O_{2}$

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3 years ago

Under what conditions of temperature and pressure do real gases behave most like ideal gases?

bekas [8.4K]

Answer:

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them

Explanation:

answer right in there:)

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3 years ago

Which of the following is an indicator of a chemical reaction?

Gemiola [76]

Answer:

option . D

Increase in temperature

is an indicator of a chemical reaction

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3 years ago

Read 2 more answers

1) What is the mass of 6.2 mol of K2CO3?

lorasvet [3.4K]

Answer:

$\boxed {\boxed {\sf 856.8648 \ grams \ of \ K_2CO_3}}$

Explanation:

To convert from moles to grams, we must find the molar mass.

1. Molar Mass

First, identify the elements in the compound. K₂CO₃ It has potassium, carbon, and oxygen. Find these elements and their masses on the Periodic Table.

K: 39.098 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Note the subscript of 2 after K and 3 after O. We must multiply oxygen's molar mass by 2, then oxygen's by 3, and add carbon.

2(39.098 g/mol) + 3(15.999 g/mol) + 12.011 g/mol= 138.204 g/mol

2. Convert Moles to Grams

Use the molar mass as a fraction.

$\frac {138.204 \ g \ K_2CO_3}{1 \ mol \ K_23CO_3}$

Multiply by the given number of moles: 6.2

$6.2 \ mol \ K_2CO_3 *\frac {138.204 \ g \ K_2CO_3}{1 \ mol \ K_23CO_3}$

$6.2 *\frac {138.204 \ g \ K_2CO_3}{1 }$

$6.2 * {138.204 \ g \ K_2CO_3}$

$856.8648 \ g \ K_2CO_3$

There are <u>856.8648 grams</u> of potassium carbonate in 6.2 moles.

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Exothermic gives off heat/energy and endothermic takes in heat/energy. Exothermic example: a candle flame

Endothermic example: baking bread

In Exothermic, you can expect the surrounding temp. to rise, and in Endothermic you can expect the surrounding temperature to fall.

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3 years ago