Answer:
The specific heat of zinc is 0.361 J/g°C
Explanation:
<u>Step 1:</u> Data given
44.0 J needed
Mass of solid zinc = 10.6 grams
Initial temperature = 24.9 °C
Final temperature = 36.4 °C
<u>Step 2</u>: Calculate the specific heat of zinc
Q = m*c*ΔT
⇒ with Q = heat (in Joule) = 44.0 J
⇒ with m = the mass of the solid zinc = 10.6 grams
⇒ with c = the specific heat of the zinc = TO BE DETERMINED
⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C
44.0 J = 10.6 grams * c * 11.5°C
c = 44.0 J / (10.6g * 11.5 °C)
c = 0.361 J/g°C
The specific heat of zinc is 0.361 J/g°C
Answer: 6.02214076 atoms Ca
Explanation:
Ca is monoatomic, so atoms in 1 mol = avogadro number
Answer:
yes
Explanation:
Solubility is an observation and no chemical reaction takes place. The composition of the compound/element is not changed.
- Hope that helped! Please let me know if you need further explanation.
Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
The answer would be A from what I know
