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mr_godi [17]
3 years ago
7

A bubble of helium gas has a volume of 0.65 L near the bottom of a large

Chemistry
1 answer:
Flura [38]3 years ago
8 0

Answer:

0.73L

Explanation:

The following data were obtained from the question :

V1 = 0.65 L

P1 = 3.4 atm

T1 = 19°C = 19 + 273 = 292K

V2 =?

P2 = 3.2 atm

T2 = 36°C = 36 + 273 = 309K

The bubble's volume near the top can be obtain as follows:

P1V1 /T1 = P2V2 /T2

3.4 x 0.65/292 = 3.2 x V2 /309

Cross multiply to express in linear form as shown below:

292 x 3.2 x V2 = 3.4 x 0.65 x 309

Divide both side by 292 x 3.2

V2 = (3.4 x 0.65 x 309) /(292 x 3.2)

V2 = 0.73L

Therefore, the bubble's volume near the top is 0.73L

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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 ofBaSO_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

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NH3 has 1 Nitrogen atom and 3 hydrogen atoms.

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