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AnnyKZ [126]
3 years ago
14

if i add water to 100ml of a 0.15 m naoh solution until the final volume is 150 ml,what will the molarity of the diluted solutio

n be?
Chemistry
2 answers:
Katena32 [7]3 years ago
7 0

Answer:

0.1 mol×L

Explanation:

Concentration= MOLES of SOLUTE / Volume of SOLUTION

So all we need to is to calculate the one quantity; Volume of SOLUTION has been specified to be 150 ml

So, MOLES of SOLUTE

x 100 x 10-L = ??mol; this was our starting solution.

And final CONCENTRATION=

0.15 . mol. 5 x 100 x 10-3L/

= 0.15 mol · L-1 150 x 10 3

=0.1 mol · L

Kay [80]3 years ago
7 0

Answer:

0.1 moles

Explanation:

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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

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Complete combustion of a sample of a hydrocarbon on excess oxygen produces equimolar quantities of carbon dioxide and water. Whi
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) determine the henry's law constant for ammonia in water at 25°c if an ammonia pressure of 0.022 atm produces a solution with a
Nataly [62]

Answer:

a. 59 m/atm

Explanation:

  • To solve this problem, we must mention Henry's law.
  • <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
  • It can be expressed as: C = KP,

C is the concentration of the solution (C = 1.3 M).

P is the partial pressure of the gas above the solution (P = 0.022 atm).

K is the Henry's law constant (K = ??? M/atm),

∵ C = KP.

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3 years ago
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