Answer:
2.122×10^25atoms
Explanation:
number of moles=mass/molar mass
7.05moles= mass of pyridine/79
reacting mass of pyridine=556.95
C5H5N= (12×5)+(5)+(14)=79
C5=60
to find the mass of carbon in 556.95g of pyridine we take the stoichometric ratio
60[C5] -----> 79[C5H5N]
x[C5] --------> 556.95g[C5H5N]
cross multiply
x=(60×556.95)/79
x=423g of carbon
moles=mass/molar mass
moles of carbon=423/12
moles=35.25moles of carbon
moles=number of particles/Avogadro's constant
35.25=number of particles/6.02×10^23
number of particles=2.122×10^25atoms of carbon
Answer: number of atoms is 5.21 · 10^24
Explanation: Atomic mass of Be is 9.012 g/mol.
Number of moles n = m/M = 78.0 g / 9.012 g/mol =
Multiply this with Avogadro number Na = 6.022*10^23 1/mol
Answer:
partial pressure of gas D Pd = 15.5 kPa
Explanation:
As per the Dalton's law of partial pressure, in a mixture, pressure exerted by each gas when summed gives the total partial pressure exerted by mixture.
P(Total) = P1+P2+P3.....
Given P(Total) = 35.7 kPa
Partial pressure of gas A Pa = 7.8 kPa
Partial pressure of gas B Pb = 3.7 kPa
Partial pressure of gas C Pc = 8.7 kPa
There, Partial pressure of gas D Pd = P(Total) -(Pa+Pb+Pc)
Pd = 35.7-(7.8+3.7+8.7) = 35.7-20.2 kPa = 15.5 kPa
Therefore, partial pressure of gas D Pd = 15.5 kPa
To find them you would have numbers of the elements in percentage or grams then you divide them by their molar mass to get their moles. From there you divide by the smallest number. Round it to two or one sig fig. If you have a number that is for ex. 2.5 you multiply it by 2 to make it whole as well the other whole numbers. Then to find the molecular formula the problem must give you another molar mass and using your empirical formula convert it to its molar mass then you divide them, larger number over smaller number. You should get a number round it to 1 sig fig. Now you use that number and multiply the subscripts on the empirical formula to get the molecular formula.
Answer:
The first thing that you need to do here is to figure out the mass of the sample.
To do that, you can use its volume and the fact that aluminium is said to have a density of
2.702 g cm
−
3
, which implies that every
1 cm
3
of aluminium has a mass of
2.702 g
.
Explanation:
