Answer:
A)P5010
Explanation:
Penta- means 5 and deca/deco- means 10
Answer:
1) 90.0 mL
2) 11.25 M
3) 0.477 M
4) 144 mL
Explanation:
The main formula that will be used for all these calculations is:
C₁V₁ = C₂V₂
C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.
For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.
1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:
5.00 M · V₁ = 500.0mL · 0.900 M - divide by 5.00
C₁ = 90.0 mL
2) This time we're finding the initial concentration:
20.0mL · C₁ = 150.0mL · 1.50 M - divide by 20.0mL
C₂ = 11.25 M
3) Now we're finding the final concentration:
12.00mL · 3.50 M = 88.0mL · C₂ - divide by 88.0mL
C₂ = 0.477 M
4) Finally, we're looking for the final volume:
9.0mL · 8.0 M = 0.50 M · V₂ - divide by 0.50 M
V₂ = 144mL
Answer:
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
Explanation:
Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.
Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
Now, we have to balance Li by multiplying LiNO₂ by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
The equation is now balanced.
Answer:
2Li + F₂ → 2LiF
Explanation:
The reaction expression is given as:
Li + F₂ → LiF
We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.
Let use a mathematical approach to solve this problem;
Assign variables a,b and c as the coefficients that will balance the expression:
aLi + bF₂ → cLiF
Conserving Li: a = c
F: 2b = c
let a = 1, c = 1 and b =
Multiply through by 2;
a = 2, b = 1 and c = 2
2Li + F₂ → 2LiF