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KiRa [710]
3 years ago
8

You have two cups one is 6cm and other 10cm with no measurement and you have to pour 8cm of water into the 10cm cup using the tw

o cup, without measuring it. How can you?
Chemistry
1 answer:
Korolek [52]3 years ago
5 0

Here’s one way to do it.

1. Fill the 6 cm cup.

2. Pour its contents into the 10 cm cup. This leaves 4 cm yet to be filled.

3. Refill the 6 cm cup and use it to fill the 10 cm cup. This leaves 2 cm in the 6 cm cup.

4. Empty the 10 cm cup and add the 2 cm from the 6 cm cup.

5. Refill the 6 cm cup.

6. Pour its contents into the 10 cm cup.

The 10 cm cup now contains 8 cm of water.

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You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40,
eimsori [14]
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95 
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
          = 7.55 - 1.95 = 5.6 
5 0
3 years ago
At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO
erastovalidia [21]

Answer : The concentration of NO at equilibrium is 0.9332 M

Solution :  Given,

Concentration of N_2 and O_2 at equilibrium = 0.200 M

Concentration of N_2 and O_2 at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}

K_c=6.25

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

                         N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially             0.200   0.200         0

.800

At equilibrium  (0.200-x) (0.200-x)  (0.800+2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}

By solving the term x, we get

x=2.6\text{ and }-0.0666

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of NO at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of NO at equilibrium is 0.9332 M

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3 years ago
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If the ion is a cation, it has a positive charge because it LOST electrons. If its an anion, then it has a negative electron because it GAINED electrons. 
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Answer:

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Explanation:

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Wat the options that give u
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