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2 years ago

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A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

joules, what is the approximate final temperature of the water? 75 °C 78 °C 81 °C 87 °C

1 answer:

Sedbober [7]2 years ago

7 0

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

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What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c

brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

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Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

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$c=1.4\times 10^{-4}$

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How many acidic protons are there in 0.6137 g of KHP?

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0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

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Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now, 1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

Learn more: brainly.com/question/16672114

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