The polyatomic nitrate ion (NO3−) is present in both the products and the reactants of the chemical equation shown below.
2 answers:
Answer:
There is one nitrate ion in the products, and two in the reactants of this equation.
Explanation:
1: We know a nitrate ion has the formula , so we just need to count how many of them are on each side of the equation.
2: To find how many are in the reactants of the equation, you look at the left hand side (before the arrow). You can see the section , which shows that there are two nitrate ions in the reactant side (as seen by the little 2).
3: To find how many are on the products side, you do the same thing - this time there is only one in the lithium nitrate.
So, there is one nitrate ion in the products, and two in the reactants of this equation.
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Answer:
The correct answer is option B.
Explanation:
Moles of = 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles
Then ,48 moles of NaOH will reacts with:
of
Then ,40 moles of will reacts with:
of NaOH
As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.
Moles left after reaction = 40 mol - 32 mol = 8 mol
Hence, the is an excessive reagent.
Answer: T2= 962.2 K
Explanation:
The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.
So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so
V1=V2 ⇒