When the balanced reaction equation of methane combustion is:
CH4 + 2O2 →CO2 + 2H2O
so, we can see that each 1 mole of methane combusted will give 2 moles of water as a product.
so first, we need to get the moles of methane =
= mass of methane /molar mass of methane
= 52.6 g / 16.04 g/mol
= 3.28 moles
when 1 mol of methane produces→ 2 moles of water
∴ 3.28 moles methane produces → X moles of water
∴ moles of water = 3.28 * 2
= 6.56 moles
when each 1 mole of water has 6.02 x 10^23 (Avogadro's number ) individual molecules:
∴number of molecules of water = 6.56 * 6.02 x 10^23
= 3.9 x 10^24 molecules
On the edges or the outer magnetic fields
Maybe. Like in hitch hikers guide to the galaxy, I need to consult with the super computer
The statement which is true about <em>Atom A</em> and <em>Atom B</em> is; Choice B: <em>Atom B will give up electrons to form bonds.</em>
According to the question;
For Atom A:
- the electron configuration is; <em>1s²2s²2p³</em>
- Therefore, it has 5 valence electrons.
For atom B:
- the electron configuration is; <em>1s²2s²2p⁶3s²3p¹</em>
- Therefore, it has 3 valence electrons.
Since, electrons need 8 electrons to assume a full octet;
Consequently, Atom B will give up electrons to become stable.
Read more:
brainly.com/question/21400937
Answer:
Explanation:
To find the theoretical yield of the equation. First identify the limiting reactant in a chemical equation.
Step 1: write out the equation and balance it.
Al+ 3mno2=3mn+ 2Alo3.
The limiting reactant is mn02 because it is not found in excess.
Step 2: convert the % to gram . All contain 67.2% mole and mno2 will be 100-67.2= 32.8
All=67.2÷100×290(total gram of the reactants)=194.88g
Mno2=32.8÷100×290g=94.12g.
Step 3:calculate the molar mass of mno2 and that of mn. The atomic mass of mn is 54.9380 and that of oxygen is 16.
Mno2=54.938+ (16)2=86.98g/mol.
Mn=54.938.
Step 4:
From your balanced equation , calculate mn.
94.12g mno2× (1mol mno2÷86.98(molarmass) of mno2×3 mol of mn/4molAl×54.938g of mn÷1mol of mn.
94.12g×1÷86.98g×3÷4×54.938÷1
=44.58g
