2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
A. 1.01 is the right answer
Since
The formula is Pv= nRT
P=1 atm
V= 22.4 L
N= x
r= 0.0821
t = 273 k (bc it’s standard temperature)
So (1)(22.4)=(x)(0.0821)(273)
X= 1.001
Answer:
- <u>259,000 g of chalk.</u>
Explanation:
<u>1) Data:</u>
a) 2000 boxes
b) 175 g / box
c) % yield = 74%
<u>2) Formula: </u>
- % yield = (theoretical yield / actual yield) × 100
<u>3) Solution:</u>
a) Calcualte the actual yield:
- mass of product = 2000 box × 175 g/ box = 350,000 g
b) Solve for the theoretical yield from the % yield formula:
- % yield = (theoretical yield / actual yield) × 100
⇒ theoretical yield = % yield × actual yield / 100
theoretical yield = 74% × 350,000g / 100 = 259,000 g
Looking at this equation P= (pa*pb)/ (pa+(pb-pa)) ya where pa=vap press a and ya= vap composition a and P= total pressure,it relates vapor pressure mixture to vapor composition. This is derived using the combination of Dalton's and Raoult's laws.
Answer:
Option A
Explanation:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion
When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-
As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible
Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-
And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-
∴ Increasing order of concentration will be PO43- < NO3- < Na+
