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jek_recluse [69]
4 years ago
8

A dioxin-contaminated water source contains 0.083% dioxin by mass. How much dioxin is present in 3.3 L of this water? Assume a d

ensity of 1.00 g/mL
Chemistry
1 answer:
KatRina [158]4 years ago
3 0

Answer:

2.739 grams of dioxin is present in 3.3 L of this water.

Explanation:

Volume of water , V = 3.3 L = 3.3 × 1000 mL = (1 L = 1000 mL)

Mass of water = m

Density of water  =d = 1.00 g/mL

d=\frac{m}{V}

m=d\times V=1.00 g/mL\times 3.3\times 1000 mL=3300 g

Percentage of dioxon in water = 0.083%

Let the mass of dioxon present in 3300 grams of water be = x

0.083\%=\frac{x}{3300 g}\times 100

x=\frac{0.083}{100}\times 3300 g=2.739 g

2.739 grams of dioxin is present in 3.3 L of this water.

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Answer:

When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral.

I hope it helps.

Have a great day.

7 0
3 years ago
Read 2 more answers
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate th
Andreas93 [3]

Complete question:

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.

Answer:

the heat of combustion of the methanol is 402.31 kJ/mol

Explanation:

Given;

mass of water, m_w = 100 g

initial temperature of water, t₁ = 28 ⁰C

final temperature of water, t₂ = 58 ⁰C

specific heat capacity of water = 4.184 J/g⁰C

reacting mass of the methanol, m = 1.00 g

molecular mass of methanol = 32.04 g/mol

number of moles = 1 / 32.04

                             = 0.0312 mol

Apply the principle of conservation of energy;

n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} =  mc\Delta t\\\\n\Delta H_{methanol} =  100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} =  12,552 \ J\\\\n\Delta H_{methanol} =  12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol

Therefore, the heat of combustion of the methanol is 402.31 kJ/mol

                       

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