When 1 molecule of calcium phosphate is added to water, what and how many ions are formed? remember that ions are charged and so
you must indicate the correct charge on each ion to receive credit for this question?
1 answer:
Asnwer :When 1 molecule of calcium phosphate is added to water, total 5 ions are formed 3 ions of Calcium() and 2 ions of Phosphate(
)
Explanation : When Calcium phosphate is added to water it gets dissociated into ions. Calcium phosphate () dissociates into 3 ions of Calcium
and 2 ions of Phosphate
.
Dissociation equation of calcium phosphate is :
From the dissociation equation we can conclude that total 5 ions are formed when 1 molecule of calcium phosphate is added to water.
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Answer:
Therefore = -1835 KJ
Explanation:
Enthalpy is denoted by H.
Enthalpy: Total heat change in a chemical reaction is called enthalpy.
The change of entalpy of a reaction is denoted by
Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.
g= gas
S= solid
P₄(g)+10Cl₂(g)→ 4Cl₅(s) =?
PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) = +157KJ
P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) = -1207 KJ
If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.
We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.
PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) = -157KJ
Multiplying 4 with equation (3)
4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) =4×( -157)KJ= -628 KJ
Adding equation (2) and (4) we get
P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) =( -1207-628)KJ
⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) = - 1835KJ
⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) = -1835 KJ
Therefore = -1835 KJ
Answer:
<u>D. It will decrease by a factor of 4</u>
Explanation:
According to the question , the equation follows :
Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.
.................(1)
STEP": First, find out the power "a" and "b"
a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)
According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>
r' = 2r if [A'] = 2[A]
Here [B] is uneffected means [B']=[B]
hence new rate =
Put the value of [A'] , [B'] and r' in the above equation:
...........(2)
Divide equation (1) by (2) we , get
Here A and A cancel each other
B and B cancel each other
We get,
1^b = 1 ( power of 1 = 1)
This is possible only when a = 1
We know that : a + b = 3
1 + b = 3
b =3 -1 = 2
b = 2
Hence the rate law becomes :
<u>.............(3)</u>
Look in the question now, it is asked to calculate the concentration of [B],if cut in half
Hence
[B']=1/2[B]
Insert the value of [B'] in equation (3)
............(a)
But
..............(b)
Compare equation (a) and (b) , we get
new rate r' =
<u>r' = 1/4 r</u>