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3 years ago

Which hydrocarbon can have cis-trans isomers?

Chemistry

1 answer:

Blizzard [7]3 years ago

3 0

Answer:

Alkenes of the type R–CH=CH–R can exist as cis and trans isomers;

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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago

A solution 0.20 molar in monomer (styrene) and 4.0 X 10-3 M in benzoyl peroxide initiator is heated at 60°C. kp = 145 liter/mole

Elden [556K]

Answer:

Part A and B are attached

Explanation:

4 0

3 years ago

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Which of these is an example of matter?

wolverine [178]

Answer:

Air

Explanation:

It takes up space/ the rest do not

7 0

3 years ago

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Stemscopedia

Alexeev081 [22]

What’s the question

7 0

3 years ago

Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha

Aleks04 [339]

Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

$pH = pKa + log \frac{[Base]}{[Acid]}$

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

$7.5 = 6.3 + log \frac{[base]}{[acid]}$

$1.2 = log \frac{[base]}{[acid]}$

$\frac{[Base]}{[Acid]} = 10^{1.2}$

$\frac{[Base]}{[Acid]} = 15.8$

$[Base] = 15.8 \times [Acid]$

The total of mole fraction of acid and base is 1. Therefore we have,

$[Acid] + [Base] = 1$

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

$[Acid] + 15.8 \times [Acid] = 1$

$16.8 [Acid] = 1$

$[Acid] = \frac{1}{16.8}$

$[Acid] = 0.0595$

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

4 0

4 years ago

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