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BabaBlast [244]

1 year ago

8

A semiconductor:

1 answer:

Alex1 year ago

7 0

Answer:

D. has characteristics between conductor and insulator

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How many total atoms are in the compound:<br> C6H12O6<br><br> 18<br><br> 6<br><br> 12<br><br> 24

Mice21 [21]

D.) There are 24 atoms in total in that compound.....

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What mineral is this? Egg-sized piece. Transparent pink color with glassy luster. Shows conchoidal fracture. The hardness is 7.

fgiga [73]

ruby

Explanation:

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Cómo podemos aplicar los conocimientos químicos a nuestra vida cotidiana?

lianna [129]

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7 0

2 years ago

A and B are powders, A is insoluble while B dissolves to give a pH 3 solution. Mixing A and B gives bubbles or effervescence and

Vladimir [108]

Answer:

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6 0

2 years ago

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5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas

Artist 52 [7]

Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

$\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)$

where,

$\Delta U$ = change in internal energy = ?

n = number of moles of gas = 5 moles

$C_v$ = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298K$

$T_2$ = final temperature = $300^oC=273+300=573K$

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

$\Delta U=nC_v(T_2-T_1)$

$\Delta U=(5moles)\times (2R)\times (573-298)$

$\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)$

$\Delta U=22863.5J=22.86kJ$

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

$\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)$

where,

$\Delta H$ = change in enthalpy = ?

n = number of moles of gas = 5 moles

$C_p$ = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298K$

$T_2$ = final temperature = $300^oC=273+300=573K$

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

$\Delta H=nC_p(T_2-T_1)$

$\Delta H=(5moles)\times (3R)\times (573-298)$

$\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)$

$\Delta H=34295.25J=34.29kJ$

The change in enthalpy of the gas is 34.29 kJ.

4 0

2 years ago