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.
Answer:
26.73 mg.
Explanation:
- Firstly, we can calculate the no. of moles of magnesium chlorate (Mg(ClO₃)₂):
no. of moles of magnesium chlorate (Mg(ClO₃)₂) = mass/molar mass = (72.03 mg)/(191.21 g/mol) = 0.377 mmol.
<em>Every 1.0 mole of magnesium chlorate (Mg(ClO₃)₂) contains 2.0 moles of Cl.</em>
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∴ The no. of moles of Cl in magnesium chlorate (Mg(ClO₃)₂) = 2(0.377 mmol) = 0.754 mmol.
∴ The mass of Cl are found in 72.03 mg of magnesium chlorate (Mg(ClO₃)₂) = (no. of moles of Cl)(atomic mass of Cl) = (0.754 mmol)(35.453 g/mol) = 26.73 mg.
Answer: 1
Explanation: coarse adjustment