Answer:
Thermal energy in the form of heat moved from the fire to the skewer and marshmallow.
Explanation:
He holds a marshmallow on a metal skewer over a fire. A few seconds late, the marshmallow is cooked and the skewer feels warm.
Answer:
54.7°C is the new temperature
Explanation:
We combine the Ideal Gases Law equation to solve this.
P . V = n. R. T
As moles the balloon does not change and R is a constant, we can think this relation between the two situations:
P₁ . V₁ / T₁ = P₂ . V₂ / T₂
T° is absolute temperature (T°C + 273)
68.7°C + 273 = 341.7K
(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂
1.63 atm.L/K = 532.5 atm.L / T₂
T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K
T° in C = T°K - 273 → 326.7K + 273 = 54.7°C
<span>Henry divides 1.060 g by 1.0 mL to find the density of his water sample.
</span>He should include THREE significant figures in the density value that hereports.
1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.