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3 years ago

Homework . Answered

Chemistry

1 answer:

bija089 [108]3 years ago

7 0

We have that the name (not chemical symbol) of the main group element in period 5 and group 3A is

Metalloid boron (B)

From the Question we are told that

The element belongs to

Period 5 and group 3A

Generally

Group 3A of the periodic table includes the metalloid boron (B), aluminum (Al), indium (In), and thallium (Tl) gallium (Ga),

Period 5 is possessed by the metalloid boron (B) of the Group 3A

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What are the mole fraction and the mass percent of a solution made by dissolving 0.21 g KBr in 0.355 L water? (d = 1.00 g/mL.) m

IRISSAK [1]

Answer:

Mol fraction H2O = 0.99991

Mol fraction KBr = 0.00009

mass % KBr = 0.059 %

mass % H2O = 99.941 %

Explanation:

Step 1: Data given

Mass of KBr = 0.21 grams

Molar mass KBr = 119 g/mol

Volume of water = 355 mL

Density of water = 1.00 g/mL

Molar mass water = 18.02 g/mol

Step 2: Calculate mass water

Mass water = 355 mL * 1g /mL

Mass water = 355 grams

Step 3: Calculate moles water

Moles water = mass water / molar mass water

Moles water = 355 grams / 18.02 g/mol

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Step 4: Calculate moles KBr

Moles KBr = 0.21 grams / 119 g/mol

Moles KBr = 0.00176 moles

Step 5: Calculate total moles

Total moles = 19.7 moles + 0.00176 moles

Total moles = 19.70176 moles

Step 6: Calculate mol fraction

Mol fraction H2O = 19.7 moles / 19.70176 moles

Mol fraction H2O = 0.99991

Step 7: Calculate mol fraction KBr

Mol fraction KBr = 0.00176 / 19.70176

Mol fraction KBr = 0.00009

Step 6: Calculate mass %

mass % KBR = (0.21 grams / (0.21 + 355) grams) *100%

mass % KBr = 0.059 %

mass % H2O = (355 grams / 355.21 grams) *100%

mass % H2O = 99.941 %

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