The correct answer is (1) one mole of NO2.
The gram formula mass is also known as the molar mass and is defined by the mass over one mole of a substance.
Hope this helps~
Answer:
A methane molecule is made from one carbon atom and four hydrogen atoms. Carbon has a mass of 12.011 u and hydrogen has a mass of 1.008 u. This means that the mass of one methane molecule is 12.011 u + (4 × 1.008u), or 16.043 u. This means that one mole of methane has a mass of 16.043 grams.
メタン分子は、1つの炭素原子と4つの水素原子から作られています。炭素の質量は12.011uで、水素の質量は1.008uです。これは、1つのメタン分子の質量が12.011 u +（4×1.008u）、つまり16.043uであることを意味します。これは、1モルのメタンの質量が16.043グラムであることを意味します。^>^
Answer:
potential energy
Explanation:
energy which is linked with the position of the object is called as potential energy. any object possessing energy due to its position is potential energy. example is holding a basketball up in to the air at a certain height so that it will have a gravitational pull towards earth surface. this gravitation pull is called as potential 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.
