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3 years ago

The concentration of C29H60 in summer rainwater is 34 ppb. Find the molarity of this compound in nanomoles per liter (nM).

Chemistry

1 answer:

Naya [18.7K]3 years ago

5 0

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter

Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.

The algebraic expression would be:

<em>ppb [=] micrograms of compound/liter of solution</em>

We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.

For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.

The respective procedure is in a attached file.

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3 years ago

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A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for

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Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

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Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

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3 years ago

A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for

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Answer:

Empirical formula is C₄H₁₀O

Explanation:

Values for C, H and O are determined as centesimal composition.

64.80 g of C in 100g of compound

13.62g of H in 100 g of compound

21.58 g of O in 100 g of compound.

We convert the mass to moles:

64.80 g . 1mol/ 12g = 5.4 moles of C

13.62 g . 1 mol /1g = 13.62 moles of H

21.58 g . 1 mol/16g = 1.35 moles of O

We pick the lowest value and we divide:

5.4 moles of C / 1.35 = 4 C

13.62 moles of H / 1.35 = 10 H

1.35 moles of O / 1.35 = 1 O

Empirical formula is C₄H₁₀O, it can be the diethyl ether.

We confirm, the excersise is well done.

Molar mass = 74g/mol

74 g of compound we have (12 . 4)g of C

In 100 g of compound we may have (100 . 48) / 74 = 64.8 g

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3 years ago

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2 years ago