I believe that the balanced chemical reaction is:
C6H12O6 + 6 O2 → 6 CO2
+ 6 H2O
So the number of grams
of oxygen required is:
mass O2 required = 48
g C6H12O6 * (1 mole C6H12O6 / 180.16 g) * (6 mole O2 / 1 mole C6H12O6) * (32
grams O2 / 1 mole)
<span>mass O2 required =
51.15 grams</span>
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)
Using the standard enthalpies of formation given in the source below:
(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
Answer:
m = 0.531 molal
Explanation:
∴ m fructose = 3.35 g
∴ V water = 35.0 mL
∴ ρ H2O = 1 g/mL
- molality = moles solute / Kg solvent
∴ Mw fructose = 180.16 g/mol
⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose
⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O
⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O
⇒ m = 0.531 molal
Answer:
The correct options are: Z1) They tend to form cations, and Z4) They tend to form ionic compounds when they combine with the elements of Group VIIA.
Explanation:
The chemical elements that have a tendency of <u>losing their valence electrons</u> and are generally present on the <u>left-side of the periodic table</u> are known as metals.
The general characteristics of metals are malleability, ductility, lustre, high thermal and electrical conductivity.
The metals<u> readily lose their valence electrons</u> present in their valence electron shell to<u> form positively charged ions called </u><u>cation</u>s and thus have <u>low ionization energy</u>.
They also tend to form<u> </u><u>ionic compounds or salts</u><u> </u>with the <u>reactive non-metals belonging to the Group VIIA of the periodic table.</u>
<u>Therefore, </u><u>Z1 and Z4 </u><u>are the properties of most metals.</u>
