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3 years ago

How much heat is needed to warm 250g of water from 22°C to 98°C ?

Chemistry

1 answer:

Ksju [112]3 years ago

6 0

Answer:

79'420J

Explanation:

We know mass 250g and

∆T=98°C -22°C

=76°C

The specific heat capacity for water is 4,18Zj/g°C

THEREFORE

∆H=250g•14,8j/g°C •76°C

=79'420J

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Explanation:

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2 years ago

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

$Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)$

The half-cell reactions are:

Oxidation half reaction (anode): $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction (cathode): $Pb^{2+}+2e^-\rightarrow Pb$

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction = ?

$[Zn^{2+}]$ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

Now put all the given values in the above equation, we get:

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

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3 years ago

An ionic bond occurs between what particles

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3 years ago

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Explanation:

Since the reaction is 1st order,

Rate of reaction=∆[A]÷t

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3 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

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Answer:

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The ionization equation is

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3 years ago

How much heat is needed to warm 250g of water from 22°C to 98°C ?​

How much heat is needed to warm 250g of water from 22°C to 98°C ?