Cm^3 = mL
1.11 g/cm^3 = 1.11 g/mL
Density (g/mL) multiplied by volume (mL) will give us the mass (g)
1.11 g/mL * 387 mL = 429.57 g
And since we have 3 significant figures, we have 430. g.
Answer:
N₂O₅ + H₂O --> 2HNO₃
This reaction is a combination reaction
Explanation:
I find that making a table is an easy way to balance a table. The table would be something like this:
Reactants (left) Products (right)
N = 2 N = 1
O = 6 O = 3
H = 2 H = 1
The number next the symbol represents how many of atoms of that particular element are present.
The aim is to make the number of atoms of each element on the left side to be equal to the number of atoms of each element on the right side. As we can see, there are 1 more nitrogen than on the right, 3 more oxygen than on the right and 1 more hydrogen than on the left.
So to make the numbers of atoms of each element on right equal to the number of atoms of each element on the left, we have to add a number. This number CANNOT be a subscript number because that would change the reaction.
We can add a 2 in front of the product (this is because there are less atoms in the right side of the equation).
N₂O₅ + H₂O --> 2HNO₃
This means there are now two HNO₃ molecules so every atom in this molecule is basically multiplied by 2. So 1 nitrogen atom becomes two (1 × 2 = 2), 3 oxygen atoms become 6 (3 × 2 = 6) and 1 hydrogen atom becomes 2 (1 × 2 = 2). If we were to make a table again with the following equation - N₂O₅ + H₂O --> 2HNO₃, the table would be as so:
Reactants (left) Products (right)
N = 2 N = 2
O = 6 O = 6
H = 2 H = 2
Now the equation is balanced as we can see the number each type of atom is the same on the right and left side.
Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
<h3>
Answer:</h3>
#1. 50 g
#2. 25 g
#3. 4 half lives
<h3>
Explanation:</h3>
<u>We are given;</u>
- Original mass of a radioisotope as 100 g
- Half life of the radioisotope as 10 years
We need to answer the questions:
#a. Mass remaining after 10 years
Using the formula;
Remaining mass = Original mass × 0.5^n , where n is the number of lives.
In this case, since the half life is 10 years then n is 1
Therefore;
Remaining mass = 100 g × 0.5^1
= 50 g
Therefore, 50 g of the isotope will remain after 10 years
#b. Mass of the isotope that will remain after 20 years
Remember the formula;
Remaining mass = Original mass × 0.5^n
n = Time ÷ half life
n = 20 years ÷ 10 years
= 2
Therefore;
Remaining mass = 100 g × 0.5^2
= 25 g
Thus, 25 g of the isotope will be left after 20 years
#3. Number of half lives in 40 years
1 half life = 10 years
But; n = time ÷ half life
= 40 years ÷ 10 years
= 4
Thus, the number of half lives in 40 years is 4.
