Answer:
47.5 g of water can be formed
Explanation:
This is the reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O
Methane combustion.
In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.
As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.
1.320 mol .2 = 2.64 moles
Now, we can convert the moles to mass (mol . molar mass)
2.64 mol . 18g/mol = 47.5 g
mass H2O = 47.56 g
balanced reation:
∴ moles CH4 = 1.320 mol
⇒ moles H2O = (1.320 mol CH4)×(2 mol H2O/mol CH4)
⇒ moles H2O = 2.64 mol H2O
∴ molar mass (mm) H2O = 18.015 g/mol
⇒ mass H2O = (2.64 mol H2O)×(18.015 g/mol)
⇒ mass H2O = 47.56 g
The phase's composition is as follows: 27
What is the alloy's composition?
Both have mass fractions of w a = w b = 0.5.
A-B alloy composition; C o = 57 wt% B - 43 wt% A
C b = 87 wt% B - 13 wt% A is the new phase composition.
Using the Lever rule, we can calculate the mole fraction (x i) or mass fraction (w i) of each phase of a binary equilibrium phase as follows:
W a = W b = 0.5
This provides us with;
0.5 = (C b - C o)/(C b - C a)
(87 - 57)/(87 - C a) = 0.5
30/0.5 = 87 - C a
60 = 87 - C a
C a = 87 - 60
C a = 27
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See explanation
1) Group 6A elements include; O, S, Se, Te, Po
Na2O, Na2S, Na2Se, Na2Te, Na2Po
2) Group 7A elements include; F, Cl, Br, I, At
AlF3, AlCl3, AlBr3, AlI3, AlAt3
3) Group 5A elements are;
N, P, As, Sb,Bi
Mg3N2, Mg3P2, Mg3As2, Mg3Sb2, Mg3Bi2
I completely agree with the first person's response
3.6 × 10^24 molecules
Concept: Moles and Avogadro's number
In this case;
We are given 6 moles of Methane;
But 1 mole of methane = 6.022 × 10^23 molecules
Therefore;
Molecules = Number of moles × Avogadro's number
= 6 moles × 6.022 × 10^23 molecules/mole
= 3.613 × 10^24 molecules
= 3.6× 10^24 molecules
Thus, there are 3.6× 10^24 molecules in 6 moles of methane.
