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1 year ago

If a 300.0 ml sample of a gas is heated at constant pressure from 25.0ºc to 55.0ºc, its new volume is _____ ml?

Chemistry

1 answer:

jekas [21]1 year ago

8 0

The new volume is 330.2 ml

calculation

The new volume is calculated using the Charles law formula

that is V1/T1= V2/T2

where T1= 25.0 c into kelvin = 25 +273 = 298 K

V1= 300.0 ml

T2 = 55.0 c into kelvin = 273 +55 =328 K

V2 = ? ml

make V2 the subject of the formula by multiplying both side by T2

V2= V1T2/ T1

V2 =[ (300.0 ml x 328 k) / 298 k} = 330.2 ml

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Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a

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Answer : The correct option is, $30.9^oC$

Explanation :

Formula used :

$q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

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Now put all the given values in the above formula, we get the final temperature of the calorimeter.

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$24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC$

$T_{final}=30.9^oC$

Therefore, the final temperature of the calorimeter is, $30.9^oC$

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What is the number of moles in 15.0 g AsH3?

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Answer:

0.192 mol.

Explanation:

To calculate the no. of moles of a substance (n), we use the relation:

n = mass / molar mass.

mass of AsH₃ = 15.0 g.

molar mass of AsH₃ = 77.95 g/mol.

∴ The number of moles in 15.0 g AsH₃ = mass / molar mass = (15.0 g) / (77.95 g/mol) = 0.192 mol.

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1 year ago

A 10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point? for

tatyana61 [14]

When the titration of HCN with NaOH is:

HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume

= 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

= 2 mmol / 0.0998mL

= 20 mL

and according to the molar ratio so, moles of CN- = 2 mmol

∴the molarity of CN- = moles / total volume

= 2 mmol / (10mL + 20mL ) = 0.0662 M

when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

= 14 - 9.31 = 4.69

when Pkb = -㏒Kb

4.69 = -㏒ Kb

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq)

by using the ICE table:

∴ the initials concentration are:

[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

[HCN] = [OH-]= X

when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

= -㏒ 0.00114

POH = 2.94

∴PH = 14 - 2.94 = 11.06

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