Can someone please help me with these three questions, it will soon be due.

1 answer:

8 0

YOU also better mark me BRAINLIEST if someone else answers!!!

volume = 100 mL

density = 1.02 g / mL

mass = density x volume

heat absorbed = 2.14 kJ

We can calculate the moles of HC1 present in 50 ml 1m solution by using the formula, moles = volume of solution in L*molarity
moles = 50/1000 = .05 , L*1 M = .05 Moles of HCL.
The same can be applied for calculation of moles fo Na OH of 50 ml, 1 M solution. (50 ml NaOH = .o5 L NaOH)
Then use the formula = volume of solution / 1000*moles.wt,
=1.825 gram.
moles=mass in grams /moles.wt= .05 moles of HCI
calculate moles of water from this reaction.
HC1+NaOH----->NACl+h20 (Balanced Equation)
.5+.5----> .5+.5
.05 moles of water
Delta H=q/n.
n=.05 to calculate the Delat H.

3) I don't understand it, sorry

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Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

$AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}$

If pH is increased then concentration of $H^{+}$ increases in solution resulting removal of $CH_{3}COO^{-}$ by forming $CH_{3}COOH$ . Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more $AgCH_{3}COO$ will dissolve
If $AgNO_{3}$ is added then concentration of $Ag^{+}$ increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less $AgCH_{3}COO$ will dissolve
Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of $Ag^{+}$ . So, more $AgCH_{3}COO$ will dissolve