Molar mass O2 = 31.99 g/mol
Molar mass CO2 = 44.01 g/mol
Moles ratio:
<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2
mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01
mass of O2 = 8174.8575 / 132.03
mass of O2 = 61.916 g
Therefore:
1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916
moles O2 = 61.916 x 1 / 31.99
moles = 61.916 / 31.99 => 1.935 moles of O2
Simply look at the periodic table and fill in what you know based on the table
The number of protons = atomic number
The number of electrons, Which is the same as the atomic number for atoms.
The number of valence electrons that is given by the group that the element is in, the top number of each column in the periodic table.
Answer:
The coefficients are 6, 1, 3
Explanation:
HNCO →C3N3(NH2)3 + CO2
From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:
6HNCO → C3N3(NH2)3 + CO2
Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:
6HNCO → C3N3(NH2)3 + 3CO2
Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.
The coefficients are 6, 1, 3
Answer : When a parallel circuit is built the voltage across each of the components remains the same, also the total current passed is the equal to sum of the currents passing through each components in the circuits.
When 2 or more components are tried to be connected in parallel they maintain the same potential difference (in voltage) across their ends of the circuit.
The potential differences across the components are the observed to be same in magnitude, and they have identical polarities between them.
Then, this same voltage is applicable to all circuit components connected in parallel.
So, if each bulb is wired to the battery in a separate loop, the bulbs will be in parallel series.
Answer:
~69.744 moles of Ca
Explanation:
Using Avogadro's constant , we know that:
1 mole = 6.022 x 10^23 atoms
S0, the number of moles in 4.20 x 10^25 atoms of Ca:
=(4.20 x 10^25 x 1 )/(6.022 x 10^23)
~69.744 moles of Ca
Q2:How many atoms are in 0.35 moles of oxygen?
1 mole = 6.022 x 10^23 atoms
S0, the number of atoms in 0.35 moles of oxygen:
=[0.35 x (6.022 x 10^23)]
=2.1077 x 10^23 atoms of Oxygen
Hope it helps:)