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Dima020 [189]
3 years ago
15

If i add water to 100 mL of a 0.15M NaOH solution until the final volume 150mL , what will the molarity of the diluted solution

be ?
Chemistry
1 answer:
labwork [276]3 years ago
3 0
0,15 moles of NaOH-------in------------1000ml
x moles of NaOH------------in--------100ml
x = 0,015 moles of NaOH

final volume = 150ml

0,015 moles of NaOH---in-------150ml
x moles of NaOH--------------in-----1000ml
x = 0,1 moles of NaOH

answer:  0,1mol/dm³  (molarity)
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Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

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The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea
sineoko [7]
We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula: 
Kc   = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 =  0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0
3 years ago
¿Que es el codigo genetico?
Burka [1]

Answer:

Los tripletes de nucleótidos de las moléculas de ADN y ARN que transportan información genética en las células vivas.

Explanation:

5 0
3 years ago
Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and
tatyana61 [14]

<u>Answer:</u> The equilibrium concentration of PCl_5 is 1.285 M.

<u>Explanation:</u>

The chemical equation for the decomposition of phosphorus pentachloride follows:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression for equilibrium constant is given as:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}

We are given:

K_c=0.042

[PCl_3]=0.18M

[Cl_2]=0.30M

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

0.042=\frac{0.18\times 0.30}{[PCl_5]}

[PCl_5]=1.285

Hence, the equilibrium concentration of PCl_5 is 1.285 M.

6 0
3 years ago
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