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2 years ago

14

A gas at constant pressure and a temperature of 293K has a volume of 8.0 L. If the temperature of the gas is increased to 314K,

what is the volume?

1 answer:

aalyn [17]2 years ago

8 0

V2 = 8.6 L

Explanation:

Given:

T1 = 293K. T2 = 314K

V1 = 8.0 L V2 = ?

Using Charles's law and solving for T2,

V1/T1 = V2/T2

V2 = (T2/T1)V1

= (314K/293K)(8.0 L)

= 8.6 L

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A collection of monomers combines to form a polymer

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3 years ago

Write the ions present in a solution of na3po4. express your answers as chemical formulas separated by a comma. offset subscript

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3 years ago

An unknown element, X, has an atomic mass of 107.868 amu. The X-109 isotope (108.905 amu) is 48.16%. What is the amu of the othe

juin [17]

Answer:

106.905 amu is the mass of the other isotope

Explanation:

The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:

X = X-109*i + X-107*i

Where X is the atomic mass = 107.868 amu

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X-107 = ?, i = 1-0.4816 = 0.5184

Replacing:

107.868amu = 108.905amu*0.4816 + X-107*0.5184

55.4194 = X-107*0.5184

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<h3>106.905 amu is the mass of the other isotope</h3>

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3 years ago

Is the equation above balanced or unbalanced? Explain? Worth 55 points.

Paul [167]

Answer:

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3 years ago

Read 2 more answers

A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

so

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

8 0

3 years ago