Answer:
The molar mass of carbon
Explanation:
Before the mass (in grams) of two moles of carbon can be determined, <u>the molar mass of the element would be needed.</u>
<em>This is because the number of mole of an element is the ratio of its mass and the molar mass</em>. That is,
number of mole = mass/molar mass
Hence, the mass of elements can be obtained by making it the subject of the formular;
mass = number of mole x molar mass
<em>Therefore, the molar mass of carbon would be needed before the mass of 2 moles of the element can be determined.</em>
Answer:
Energy in the campfire originates from the potential chemical energy of the wood, before it is burnt to warm and give light around the campfire.
Explanation:
For a camp fire, the energy input is in the form of the potential chemical energy, stored up in the firewood used to fuel the flame.
The energy output is in the form of heat energy that the campfire radiates all around, light energy given off from the flame, and a little bit of sound energy, heard in the cracking of the firewood as they burn in the flame.
chemical energy ⇒ heat energy + light energy + sound energy
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
