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3 years ago

Show the difference in the reactivity of Na,Ca,Mg,Al,Fe,Cu with H2O

Chemistry

1 answer:

Vesnalui [34]3 years ago

7 0

Answer:

Explanation:

Na react with H2O to form NAOH

2 Na+2H2O....................2NAOH + H2

Ca react with water and form calcium hydroxide

Ca + 2H2O........................Ca(OH)2

Mg react with water and form Magnesium hydroxide

Mg +2H2O .........................Mg(OH)2 however this coating of mg(oh)2 prevent it from further reaction

Fe react with water and form ferric hydride

3Fe +H2O.......................2 FeH +FeO

copper do not react with water

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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

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Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

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A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of

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Answer:

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Explanation:

Initial volume of the gas = $V_1=10.0 L$

Initial pressure of the gas = $P_1=4.0 atm$

Final volume of the gas = $V_2=5.0 L$

Final pressure of the gas = $P_2=8.0 atm$

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$PV=k$

The equation given by this law is:

$P_1V_1=P_2V_2$

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The gas in the cylinder is obeying Boyle's law.

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