C3H8 and CH4 are two compounds made from the same two elements, C and H. The ratios of C and H for both are round numbers.
Law of multiple proportions states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.
So the ans is C) Law of Multiple Proportions
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
4P + 502 -> P4O10 this is the answer
The equilibrium for the dissolution of the weak base is ;(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)
<h3>What is a weak base?</h3>
A weak base is one that does not ionize completely in solution. As such, a weak base will have a very low base dissociation constant Kb reflecting its minimal dissociation in solution.
The question is incomplete hence we are are unable to work out the equilibrium but in solution it will look like this;
(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)
Learn more about weak base: brainly.com/question/4131966
Answer:
66.2 % of O
Explanation:
Our compound is the lithium nitrite.
LiNO₂
This salt is ionic and can be dissociated: LiNO₂ → Li⁺ + NO₂⁻
We determine the molar mass:
molar mass of Li + 3 . molar mass of N + 6 . molar mass of O
6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol
The mass of oxygen contained in 1 mol of lithium nitrite is:
6 . 16 g/mol = 96 g
So the percentage of oxygen present is:
(96 g / 144.94 g) . 100 = 66.2 %
