Answer:
b. 2.28 M
Explanation:
The reaction of neutralization of NaOH with H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
<em>Where 2 moles of NaOH react per mole of H2SO4</em>
<em />
To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:
<em>Moles H2SO4:</em>
45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4
<em>Moles NaOH:</em>
0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH
<em>Molarity NaOH:</em>
0.0457moles NaOH / 0.020L =
2.28M
Right option:
<h3>b. 2.28 M</h3>
Answer:
For carbon the most important forms of hybridization are the sp2- and sp3- hybridization. Besides these structures there are more possiblities to mix dif- ferent molecular orbitals to a hybrid orbital. An important one is the sp- hybridization, where one s- and one p-orbital are mixed together.
Answer:
9 L
Explanation:
According to the question , the given reaction is -
2NO(g) + O₂(g)------->2NO₂(g)
Since ,
At STP ,
One mole of a gas occupies the volume of 22.4 L.
Hence , as given in the question -
9 L of NO , i.e .
22.4 L = 1 mol
1 L = 1 / 22.4 mol
9 L = 1 / 22.4 * 9 L = 0.40 mol
From the chemical reaction ,
The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .
Hence ,
2 moles of NO will produce 2 moles of NO₂.
Therefore ,
0.40 mol of NO will produce 0.40 mol of NO₂.
Hence , the volume of NO₂ can be calculated as -
1 mol = 22.4 L
0.40 mol = 0.40 * 22.4 L = 9 L
Answer:
2
Explanation:
Each orbital can hold two electrons. One spin-up and one spin-down.
