Answer:
See explanation
Explanation:
1) Physical change is usually reversible, while chemical change isn't
2) Chemical change involves the change of chemical composition of matter while physical change doesn't
1128 is the momentum for this. i think
Answer:
14.7°C
Explanation:
Q = m·ΔT·c
ΔT = 
ΔT =
= 1320 J / ((230 g) * (.39 J/g°C)
ΔT = 14.7 °C
Answer:
The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ
Explanation:
<u>Step 1: </u>The balanced equation
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH = -802 kJ
<u>Step 2:</u> Given data
We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.
The enthalpy change of combustion, given here as Δ
H
, tells us how much heat is either absorbed or released by the combustion of <u>one mole</u> of a substance.
In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number), 802.3 kJ of heat.
<u>Step 3: </u>calculate the enthalpy change for 3 moles
The -802 kj is the enthalpy change for 1 mole
The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ
The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ
Answer:
Freezing T° of solution = - 7.35 °C
Explanation:
This is about the freezing point depression, a colligative property which depends on solute.
The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i
Freezing T° of pure solvent is -3.1°C
At this case, i = 1. As an organic compound the urea does not ionize.
We determine the molality (mol/kg of solvent)
We convert mass to moles:
12.3 g . 1mol / 60.06 g = 0.205 moles
0.205 mol / 0.3 kg = 0.682 mol/kg
We replace data in the formula:
-3.1°C - Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1
Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C
Freezing T° of solution = - 7.35 °C
