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3 years ago

Balance this equation. Not all spots have to have a number.

Chemistry

1 answer:

Lisa [10]3 years ago

3 0

2Na+2H2O----2NaOH+H2

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Why are plastic containers preferred to glass containers for storing chemicals in the laboratory?

Nookie1986 [14]

Plastics are non-corrosive and non-reactive in nature. So they are used for storing chemicals in the laboratory. They are used for strong chemicals because they do not react with chemicals neither do they corrode

7 0

4 years ago

If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta

blondinia [14]

Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH $_{2}$ ) $_{3}$ CNH $_{2}$

if TRIS is react with HCL it will form salt

(HOCH $_{2}$ ) $_{3}$ CNH $_{2}$ + HCL ⇆ (HOCH $_{2}$ ) $_{3}$ NH $_{3}$ CL

Let the reference volume is 100

Mole of TRIS is = 100 × 0.2 = 20

Mole of HCL is = 100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk $_{a}$ + log (base/ acid)

8.3 + log(10/10)

8.3

6 0

3 years ago

PLEASE ANSWER ASAP! DOUBLE CHECK THE ANSWER!

hammer [34]

I believe the answer would be C, unbalanced.
Since the forces are in opposite directions, and aren't the same size, the answer would be unbalanced.

7 0

3 years ago

All of the following require breaking and forming chemical bonds except

skad [1K]

As far as I know, the answer is B (breaking a rock)

3 0

3 years ago

A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio

Pachacha [2.7K]

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ( $K_{a}$ ) of HA is represented as-

$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$

6 0

3 years ago