The choices can be found elsewhere and as follows:
a. mass-mass problems
<span>b. mass-volume problems </span>
<span>c. mass-particle problems </span>
<span>d. volume-volume problems
</span>
I believe the correct answer is option D. It is volume-volume problems that does not require the use of molar mass. <span> Here you are dealing with molarities and volumes to determine concentrations. Molar mass is not part of any calculations.</span>
Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is 
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- 
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
Answer:
The pressure increases to 3.5 atm.
Solution:
According to Gay-Lussac's Law, " At constant volume and mass the pressure of gas is directly proportional to the applied temperature".
For initial and final states of a gas the equation is,
P₁ / T₁ = P₂ / T₂
Solving for P₂,
P₂ = P₁ T₂ / T₁ ----- (1)
Data Given;
P₁ = 3 atm
T₁ = 27 °C + 273 = 300 K
T₂ = 77 °C + 273 = 350 K
Putting values in eq. 1,
P₂ = (3 atm × 350 K) ÷ 300 K
P₂ = 3.5 atm
Atomic Numbers are the number of protons within the nucleus,
which is 58 here.
Atomic Number=58
Mass Number is the sum of the protons and neutrons in the nucleus,
=58+82
Mass Number=140
Periodic symbol of this isotope is entirely dependent on the number of protons, 58, which corresponds to the element Cerium, abbreviated as Ce in the periodic table.
Periodic Symbol=Ce
The simple equation used to calculate work is force multiplied by distance, thus as this is the case increasing the distance by a certain amount, assuming the force applied to the object is constant, the amount of work you are doing on the box for instance pushing it, is going to be greater
Since you are pushing the box with the same force covering a greater distance with the force.
