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Answer: 1.09 g
Explanation:
If we use the approximation that 1 mole is 22.4 L, then setting up a proportion,
- 1/22.4 = x/0.345 (x is the number of moles in the sample)
- x = 0.0154 mol
Since the mass of a mole of chlroine is about 70.9 g/mol, (0.0154)(70.9) = 1.09 g (to 3 s.f.)
Given:
m = 52 kg. the mass of water
ΔT = 22 °C, the temperature increase
Note that
c = 4.184 J/(g-°C), the specific heat of water.
The quantity of heat required is
Q = mcΔT
= (52,000 g)*(4.184 J/(g-°C))*(22 °C)
= 4.7865 x 10⁴ J
Because 1 J = 2.39 x 10⁻⁴ cal. the heat required is
Q = (4.7865 x 10⁴ J)*(2.39 x 10⁻⁴ cal/J) = 1143.97 cal ≈ 1144 cal
Answer: 1144 cal
The answer is (3) An electron in the third shell has more energy than an electron in the second shell. The energy of electron will increase when number of shell increase.
Graphite is a polymorph of the element carbon.