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Naddik [55]
4 years ago
10

What happens to the heat energy which is supplied to the solid once it starts melting?

Chemistry
1 answer:
AleksAgata [21]4 years ago
5 0
The heat energy is transferred to the solid substance
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The lithospheric plates float on top of the _______, which is a weak part of the mantle that flows slowly.
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4 years ago
The combustion of 1.685 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00 K to 302.16 K. The he
AfilCa [17]

Answer:

ΔH =  - 2020.57 kJ/mol

Explanation:

Given that :

mass of propanol = 1.685 g

the molar molar mass = 60 g/mol

Thus; the number of  moles = mass/molar mass

= 1.685 g/60 g/mol

= 0.028 g/mol

However ;

ΔH = heat capacity C × Δ T

Given that:

The temperature increases from  298.00 K to 302.16 K.

Then ;

Δ T = 302.16 K - 298.00 K

Δ T = 4.16 K

heat capacity C = 13.60 kJ/K

∴

ΔH = 13.60 kJ/K × 4.16 K

ΔH =  56.576 kJ

The equation of the given reaction can be represented as :

C_3H_7OH_{(l)}+\dfrac{3}{2}O_{2(g)}  \to 3CO_{2(g)} +4H_2O_{(l)}

Thus for 0.028 mol of heat liberated; ΔH =  56.576 kJ

For 1 mole of heat liberated now:

ΔH =  56.576 kJ/0.028 mol

ΔH =  2020.57 kJ/mol

SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;

ΔH =  - 2020.57 kJ/mol

5 0
3 years ago
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The basic structure of an amino acid contains the amine group, NH2, and a carboxylic acid group separated by one or more carbon
OLEGan [10]

The answer would be True

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The iupac name of this ??
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3 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

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

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
4 years ago
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