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Kryger [21]
3 years ago
5

How much heat must be removed from 456 g of water at 25.0°c to change it into ice at?

Chemistry
1 answer:
N76 [4]3 years ago
5 0
This question can be simply solved by using heat formula,
    Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
 Q = 456 g x 4.186 J / g K x 25 K
     = 47720.4 J
     = 47.72 kJ

hence 47.72 kJ of heat energy should be removed.
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Element Q has an atomic number 68. Consider the isotope Q-136 how many protons are in a neutral atom of this isotope?
mote1985 [20]

Answer: 68

Explanation:

Isotopes of an element have same number of protons but different number of neutrons. Which means isotopes of an element have same atomic number but different mass number.

Atomic number is equal to the number of protons or the number of electrons for a neutral atom and is specific to a particular element.

Mass number is the sum of number of protons and the number of neutrons.

Given : atomic number of element Q = 68 = number of protons

Mass number of isotope Q-136 = 136

But as isotopes have same atomic number, the number of protons will be same and hence there are 68 protons are in a neutral atom of this isotope.

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3 years ago
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I hope this helps.  Let me know if anything is unclear.
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Learn more on (NH4)2CO3 here:brainly.com/question/2272720

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2 years ago
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