As the atomic number increases, the number of _ _, too.
2 answers:
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Answer:
103.9 g
Explanation:
- C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:
- 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈
Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:
- 1.23 mol C₃H₈ *
= 3.69 mol CO₂
We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 3.69 mol CO₂ * 44 g/mol = 162.36 g
And <u>apply the given yield</u>:
- 162.36 g * 64.0/100 = 103.9 g
Polar covalent bonds (because hydrogen and oxygen form polar bonds and are both nonmetals so it's covalent) and hydrogen bonds (because the water molecules are attracted to each other with partial charges, causing specific properties like surface tension).
So in my very bad drawing that I attached in case you're more a visual learner, the d- and d+ show the partial charges of hydrogen and oxygen (making it polar, as the electrons in the bond are more shifted towards oxygen, which is why oxygen has a negative sign) and the yellow dotted line show the hydrogen bonds.
So in my very bad drawing that I attached in case you're more a visual learner, the d- and d+ show the partial charges of hydrogen and oxygen (making it polar, as the electrons in the bond are more shifted towards oxygen, which is why oxygen has a negative sign) and the yellow dotted line show the hydrogen bonds.
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M