We will use this formlula: Mass in grams = Number of moles x Molecular mass of 1 mole.
Since, we know the avagadro number is 6.02 x 10²³, we only have two unknown values left which are the molecular mass of CH3OH and its mole.
Molecular Mass: C = 12, H= 1, O = 16, since we have C=12, H4 = 4, O = 16, we will add them up: 12 + 4 + 16 =32
We know that one mole of anything = 6.02 x 10²³.
So we will use this formula to find the mole of methanol: Number of moles = Number of molecules / Avagadro number
Number of moles of CH3OH = (9.79 x 10^24)/6.02 x 10²³) = 16.263 moles.
Now we know that the molecular mass = 32 and the mole is = 16.263.
Now we can find its mass by using this formula: <span>Mass in grams = Number of moles x Molecular mass of 1 mole.
</span>
Mass in grams = 16.263 x 32 = 520g
Answer:
0.30 M
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t = ?
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 1.36 M
k is the rate constant = 0.208 s⁻¹
t = 7.30 seconds
So,
![[A_t]=1.36\times e^{-0.208\times 7.30}\ M=0.30\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D1.36%5Ctimes%20e%5E%7B-0.208%5Ctimes%207.30%7D%5C%20M%3D0.30%5C%20M)
The solution would be like this for this specific problem:
<span>Given:
H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>
<span>
pressure H2 = 2.6 - x
pressure Cl2 = 3.14 - x
<span>pressure HBr = 2x = 1.13
x = 1.13 / 2 = 0.565
<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575
Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>
= 1.2769 / (5.240125)
= 0.24367739319195629875241525726963
= 0.244
<span>Therefore, the Kp for the reaction at the given temperature
is 0.244.
To add, </span>the hypothetical pressure of a gas if
it alone occupied the whole volume of the original mixture at the same
temperature is called the partial pressure or Kp.
Answer:
26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.
Explanation:
Molar mass of compound = 120 g/mol
Number of sulfur atom = 1
Atomic mass of sulfur = 32 g/mol
Percentage of element in compound :
Sulfur :
26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.
Answer:
The answer is 7600 nm.
Explanation:
As, Y = 0.25 = [ L ÷ (400 + (L)]
0.95 x 400 + 0.25 [ L] = [ L ]
380.25 = [ L ] - 0.95 [ L ]
= 0.05 [L]
[L] = 380 ÷ 0.05 = 7600nm
