Answer:
1 unit
Explanation:
It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units.
1/12 * 12 = 1. I think so.
Answer:
Molarity = 2.3 M
Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution
1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles
2- The volume of solution is given = 1 liter
3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M
Hope this helps :)
Answer:
Negative sign says that release of heat.
Explanation:
The expression for the calculation of the heat released or absorbed of a process is shown below as:-
Where,
is the heat released or absorbed
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass = 25.2 g
Specific heat = 0.444 J/g°C
So,
Negative sign says that release of heat.
A.
→ 
B.
→ 
C.
→ 
What is a balanced chemical equation?
An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.
A.
→ 
B.
→ 
C.
→ 
Learn more about the balanced chemical equation here:
brainly.com/question/15052184
#SPJ1
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition =
× 100
% composition =
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).