Answer:
Explanation:
To convert from moles to grams, the molar mass is used (mass of 1 mole). The values are the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units.
<h3>1. Molar Mass</h3>
We are given the compound sodium hydroxide (NaOH) and we need to look up the molar masses of the individual elements.
- Na: 22.9897693 g/mol
- O: 15.999 g/mol
- H: 1.008 g/mol
The formula for the compound has no subscripts, so there is 1 mole of each element in 1 mole of the substance. We can simply add the molar masses.
- NaOH: 22.9897693 + 15.999 + 1.008 = 39.9967693 g/mol
This means there are 39.9967693 grams of sodium hydroxide in 1 mole.
<h3>2. Convert Grams to Moles </h3>
Use the molar mass we found as a ratio.
Since we are converting 17.6 grams of NaOH to moles, we multiply by this value.
Flip the ratio so the units of grams of NaOH cancel.
<h3>3. Round </h3>
The original measurement of grams has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.
The 0 in the ten thousandths place (in bold above) tells us to leave the 0 in the thousandth place.
17.6 grams of sodium hydroxide are equal to <u>0.440 moles of sodium hydroxide.</u>
Answer: Cl2
Explanation: ionic compounds forms cations and anions in water solution.
HI is a weak acid and forms H+ and I- ions. Cl2 is dissolved slightly in water
And stays in molecule form which can not transfer charge like cations and anions.
Answer:
Cyclobutane is a cycloalkane and organic compound with the formula (CH2)4.
Explanation:
Answer:
b. 3.35
Explanation:
To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.
pH = pKa + log ([salt]/[acid]) (Eq. 01)
Where
pKa = -log(Ka) (Eq. 02)
[salt] = Molar concentration of salt produced as a result of titration
[acid] = Molar concentration of acid left in the solution after titration
Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:
HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)
This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.
Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles
Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles
As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.
Therefore
Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)
Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)
Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:
pH= -log(4.5x10 -4) + log (0.0025/0.0025)
Solving above we get
pH = 3.35
