B. The mouse is hungry , because you dont really know that only the mouse does
Answer:
A.compound genuinely bozo L XL so
Answer:
0.928 M
Explanation:
The concentration of acid can be determined by using the volume used and the concentration and volume used of base.
We will use the law of equivalence of moles.
M₁V₁=M₂V₂
M₁ = concentration of base used
V₁ = volume of base used
M₂ = concentration of acid used =? (to be determined)
V₂ = volume of acid used
The initial concentration of KOH used is diluted so let us find the final concentration of KOH after dilution
initial moles = final moles
initial concentration X initial volume = final concentration X final volume
6.2 X 2.1 = 250 X final concentration
final concentration = 0.052 M = M₁
V₁ = 36.9 mL
V₂ = 6.2 mL
Here with each mole of phosphoric acid three moles of KOH are used.
Therefore
3 M₁V₁ = M₂V₂
M₂ = 
I think that the answer is a research bc it’s just sounds like that right answer
Mass of BaO in initial mixture = 3.50g
Explanation:
Let mass of BaO in mixture be x g
mass of MgO in mixture be (6.35 - x) g
Initially CO_2
Volume = 3.50 L
Temp = 303 K
Pressure = 750 torr = 750 / 760 atm
Applying ideal gas equation
PV = nRT
n = PV / RT
(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303
(n)_CO_2 = 0.139 mole
Finally; mole of CO_2
n= PV /RT
((245/760) *3.5) / 303* 0.0821
(n)_CO_2 = 0.045 mole
Mole of CO_2 reacted = 0.139 - 0.045
=0.044 mole
BaO + CO_2 BaCO_3
Mgo + CO_2 MgCO_3
moles of CO_2 reacted = ( moles of BaO + moles of MgO)
moles of BaO in mixture = x / 153 mole
moles of MgO in mixture = 6.35 - x mole / 40
Equating,
x/ 153 +6.35/40 = 0.094
= x/153 + 6.35 / 40 - x/40 =0.094
= x (1/40 - 1153) = (6.35/40 - 0.094)
= x * 10.018464
= 0.06475
mass of BaO in mixture = 3.50g