The volume that will be occupied at 735 torr and 57 c is 23.12 L
<u><em>calculation</em></u>
- <u><em> </em></u> At STP temperature=273 k and pressure=760 torr
- <u><em> </em></u>by use of combined gas formula
that is P1V1/T1= P2V2/T2
where; P1 =760 torr
T1= 273 K
V1= 18.5 L
P2= 735 torr
T2= 57+273= 330 K
V2=?
- by making V2 the formula of subject
V2= T2P1V1/P2T1
V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L
Answer:
Na2O+H2O=2NaOH
Step by step exp.
Given:
Equation Na2O+H2O=NaOH
To find: Balance the equation
Solution:
Taking LHS of the equation
LHS=Na2O+H2O
There is 2 sodium, 2 oxygen,& 2 hydrogen
To balance the equation we have equal number of atom so we multply 2 to the RHS=2NaOH
There fore the equation form is
Na2O+H2O=2NaOH
In any state of matter there is always motion in the atoms. So the answer would be C even though solids are together and gases atoms are spread out
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Explanation:
There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.