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3 years ago

PLEASE HELP ME!!!! WILL GIVE BRAINLIEST!!!!

Chemistry

1 answer:

vitfil [10]3 years ago

4 0

Answer: 1. 3.23 m

2. 32.4

3. B adding solvent and C removing solute

Explanation:

1. Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent

$Molality=\frac{n}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent

Now put all the given values in the formula of molality, we get

$Molality=\frac{43.6mol}{13.5kg}=3.23mol/kg$

2. Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

$3.4=\frac{n}{9.54}$

$n=32.4$

Therefore, the moles of $Na_2CO_3$ is 32.4

3. Molarity can be decreased by decreasing the moles of solute and by increasing the volume of solution.

Thus adding solvent and removing solute will decrease the molarity.

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Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient

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Answer:

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Explanation:

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Reduction: $Cr_{2}O_{7}^{2-}(aq)\rightarrow Cr^{3+}(aq)$

Balance Cr: $Cr_{2}O_{7}^{2-}(aq)\rightarrow 2Cr^{3+}(aq)$
Balance O and H in acidic medium: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$
Balance charge: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$ .......(2)

$[3\times Equation-(1)]+Equation(2)$ gives balanced equation:

$3Sn^{2+}(aq)+Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 3Sn^{4+}(aq)+2Cr^{3+}(aq)+7H_{2}O(l)$

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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

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Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

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