The number of hydrogen atoms that are in 4.40 mol of ammonium sulfide is 2.12 x10^25 atoms
calculation
find the number of moles of Hydrogen in ammonium sulfide (NH4)2S
that is 4.40 x number of hydrogen atoms in (NH4)2S ( 4x2= 8 atoms)
moles is therefore= 4.40 x8= 35.2 moles
by use of Avogadro's law constant
that is 1mole = 6.02 x10^23 atoms
35.2 moles=?
by cross multiplication
{35.2 moles x 6.02 x10^23} /1 mole = 2.12 x10^25 atoms
Answer:
They do have a larger chance, but others might be infected too.
PLS GIVE BRAINLIEST
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.
Balanced equation :
1 CaCO3(s) = 1 CaO(s) + 1 CO2(g)
hope this helps!
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