Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Answer:
They have fewer hydrogen atoms attached to the carbon chain than alkanes
Explanation:
Let's compare ethane (an alkane) with ethene (an alkene) and ethyne (an alkyne):
- Ethane's formula is C₂H₄, while ethene's is C₂H₄ and ethyne's C₂H₂.
As you can see, alkenes and alkynes have fewer hydrogen atoms attached to the carbon chain due to them having multiple bonds between the carbon atoms.
1s2,2s2.2p6,3s2,3p6,3d4,4s2
Answer:
0.24M
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation above, we obtained the following information:
nA (mole of acid) = 1
nB (mole of base) = 2
Data obtained from the question include:
Va (volume of the acid) = 12mL
Ca (concentration of the acid) =?
Vb (volume of the base) = 36mL
Cb (concentration of the base) = 0.16 M
The Ca (concentration of the acid) can be obtained as follow:
CaVa/CbVb = nA/nB
Ca x 12 / 0.16 x 36 = 1 /2
Cross multiply to express in linear form as shown below:
Ca x 12 x 2 = 0.16 x 36
Divide both side by 12 x 2
Ca = 0.16 x 36/ 12 x 2
Ca = 0.24M
Therefore, the concentration of the acid is 0.24M
