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Dafna1 [17]
1 year ago
13

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]1 year 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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Answer:

indicator

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Due to the fact that a noticeable pH change occurs near the equivalence point of acid-base titrations, an indicator can be used to signal the end point of a titration.

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4 0
1 year ago
What element speed of sound is 323 m/s?​
Pepsi [2]

Answer:

Speed of Sound

Explanation:

Speed of sound, fluid phases

 

m/s

notes

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3 0
2 years ago
A 5.00 gram sample of an oxide of lead PbxOy contains 4.33g of lead. Determine the simplest formula for the compound.
Valentin [98]
PbO2

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8 0
2 years ago
Write a net ionic equation for the reaction that occurs when nickel(ii carbonate and excess hydrobromic acid (aq are combined.
seropon [69]
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Now, writing this in ionic form:
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Overall equation:
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7 0
2 years ago
What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?
dusya [7]
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

          6.86 = -㏒ Ka 

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by using ICE table:

               H2PO4- →  H+  + HPO4
initial      0.4 m            0         0

change     -X                +X       +X

Equ       (0.4-X)               X          X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X)   by solving for X

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∴PH = -㏒[H+]

        = -㏒ (2.3 x 10^-4)
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3 0
1 year ago
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