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Arlecino [84]
3 years ago
14

Please hurry Which of the following statements is true?

Chemistry
1 answer:
kykrilka [37]3 years ago
6 0
Electrons are able to move easily in an insulator
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98 elements are naturally forming elements.

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18 Reema observed that most of the fish in the pond of her village were gradually dying. She
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Answer:

a. The fish are dying due to factory pollution making the water of the pond uninhabitable.

b. The pond can be neutralized by adding something basic to balance the pH levels of the pond, or by adding significant amounts of water to dillute the acidity of the pond.

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3 years ago
What is the mass of 2.30 ×1022 molecules of NaOH
viva [34]
Molar mass of NaOH = 39.997 g/mol

39.997 g  ---------- 6.02x10²³ molecules
   g --------------- 2.30x10²² molecules

mass = 2.30x 10²² * 39.997 / 6.02x10²³

mass = 9x199^24 / 6.02x10²³

mass = 15.280 g

hope this helps!

4 0
3 years ago
6) When a substance such as ice melts, its temperature increases. Describe what happens to the arrangement of the water molecule
Vinvika [58]
When temperature increases, there is more energy in a substance. For example, when water boils, more energy is produced. The arrangement of molecules in a solid are closer together and more packed, while the arrangement of molecules in a liquid are father apart and less packed.
4 0
4 years ago
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
4 years ago
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