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tekilochka [14]
2 years ago
8

In a titration to find the concentration of 30ml of a H2SO4 solution, a student found that 40ml of 0.2M KOH solution was needed

to reach the endpoint. What's the concentration of the H2SO4?
Question 21 options:



A) 0.27M




B) 0.53M




C) 0.4M




D) 1.1M
Chemistry
1 answer:
lianna [129]2 years ago
6 0

the queation is so difficult

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They have the same amount but if you count the wrist it was more then the leg and ankle.
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Pls help me with this
Zina [86]

Answer:

[I_2]=[Br]=0.31M

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

I_2+Br_2\rightleftharpoons 2IBr

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}

Thus, we solve for x as show below:

\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M

Therefore, the concentrations of both bromine and iodine are:

[I_2]=[Br]=2.0M-1.69M=0.31M

Regards!

8 0
3 years ago
1Calculate the density of an object that has a mass of 84.7g and a volume of 59.3 cm3
Iteru [2.4K]

Answer:

<h2>1.43 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 84.7 g

volume = 59.3 cm³

We have

density =  \frac{84.7}{59.3}  \\  = 1.428330...

We have the final answer as

<h3>1.43 g/cm³</h3>

Hope this helps you

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(P O4) 2   Given
P2 O8   Distribute
We now have 2 atoms of phosphorus and 8 atoms of oxygen. 

This problem has decided to be nice and just ask for calcium, so we don't have to distribute. All you have to do for this problem is look at the subscript beside Ca, which is 3. Therefore, the correct answer is C.
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for hydrogen-like atom classify these electron transitions by whether they result in absorption or emission light n=3 to n=5 and
iren [92.7K]
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