There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16
Answer:
1.85 × 10⁻⁶
Explanation:
0.0003 ÷ 162 = 1.851851852 × 10⁻⁶ ⇒ 1.85 × 10⁻⁶
Hope that helps.
Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
Answer:
9.69g
Explanation:
To obtain the desired result, first let us calculate the number of mole of N2 in 7.744L of the gas.
1mole of a gas occupies 22.4L at stp.
Therefore, Xmol of nitrogen gas(N2) will occupy 7.744L i.e
Xmol of N2 = 7.744/22.4 = 0.346 mole
Now let us convert 0.346 mole of N2 to gram in order to obtain the desired result. This is illustrated below:
Molar Mass of N2 = 2x14 = 28g/mol
Number of mole N2 = 0.346 mole
Mass of N2 =?
Mass = number of mole x molar Mass
Mass of N2 = 0.346 x 28
Mass of N2 = 9.69g
Therefore, 7.744L of N2 contains 9.69g of N2
Answer:
Explanation:
Hello!
In this case, since the applied current for the 50.0 mins provides the following charge to the system:
As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:
Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:
Best regards!
