The answer would be A, 100g since it’s equal
Answer:
The mass of 2,50 moles of NaCl is 146, 25 g.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl= 23 g+ 35,5 g= 58, 5 g/ mol
1 mol ------ 58, 5 g
2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>
<span>Answer:
The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first.
moles HCl = 0.04000 L * 0.100 M = 0.00400 moles
moles KOH = 0.02500 L * 0.100 M = 0.00250 moles
moles HCl left = 0.00400 - 0.00250 = 0.00150 moles
Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+]
pH = -log [H+] = -log (0.0231) = 1.64</span>
The molarity of aqueous lithium bromide, LiBr solution is 0.2 M
We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.
- Volume = 10 mL = 10 / 1000 = 0.01 L
- Molarity of Pb(NO₃)₂ = 0.250 M
- Mole of Pb(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Pb(NO₃)₂ = 0.25 × 0.01
Mole of Pb(NO₃)₂ = 0.0025 mole
Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂
Pb(NO₃)₂ + 2LiBr —> PbBr₂ + 2LiNO₃
From the balanced equation above,
1 mole of Pb(NO₃)₂ reacted with 2 mole of LiBr.
Therefore,
0.0025 mole of Pb(NO₃)₂ will react with = 2 × 0.0025 = 0.005 mole of LiBr
Finally, we shall determine the molarity of the LiBr solution
- Mole = 0.005 mole
- Volume = 25 mL = 25 / 1000 = 0.025 L
- Molarity of LiBr =?
Molarity = mole / Volume
Molarity of LiBr = 0.005 / 0.025
Molarity of LiBr = 0.2 M
Learn more about molarity: brainly.com/question/10103895
Ground state means 1s which can hold 2 electrons.
l for 1s is = 0
ml = 0 (given)
possible values of m = 0
so it can hold maximum of 2 electrons. One spin up and other spin down.
