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3 years ago

The volume of oxygen gas changed from 10 ml to 11 ml as the temperature changed. is this a chemical change or a physcial change

Chemistry

1 answer:

Effectus [21]3 years ago

5 0

It's physical change, because only the volume of oxygen changed (no new substance are formed).

:•)

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what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

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3 years ago

Calculate the amount of heat required to raise the temperature of a 65-g sample of water from 32 ∘C to 65 ∘C. (The specific heat

dsp73

The amount of heat required is 9.0 kJ.

q = mCΔT

ΔT = T_f – T_i = 65 °C – 32 °C = 33 °C

q = 65 g × 4.184 J·°C⁻¹g⁻¹ × 33 °C = 9000 J = 9.0 kJ

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Answer:

81 °C

Explanation:

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