Answer:
A. Zodiac
B. Palingenesis
C. Palabra mysteria
D. Decknamen
The correct answer is D. Decknamen.
Explanation:
The balanced reaction is:<span>
Mg (s) + 2HCl (aq)
--> MgCl 2 (aq) + H 2 (g)
We
are given the amount hydrochloric acid to be used for the reaction. This will
be the starting point of the calculation.
40.0 g
HCl ( 1 mol HCl / 36.46 g HCl)
(1 mol H2 / 2 mol HCl) (2.02 g H2 / 1 mol H2) = 1.11 g H2</span>
Answer:
19.29 g.cm⁻³
Solution:
Data Given:
Mass = 301 g
Volume = 15.6 cm³
Formula Used:
Density = Mass ÷ Volume
Putting values,
Density = 301 g ÷ 15.6 cm³
Density = 19.29 g.cm⁻³
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
