Are u sure this is the right option? Well, antimony can be decomposed. Including octane.
In order to determine the amount of moles in 23g of phosphorus, you must use math.
To get the answer, you must multiply the molar mass of phosphorus, which there is 30.97g in 1 mol, and put the mol over the mass, then multiply by your desired amount you want, which is 23g; it would look like this:
(23.00g)( <u>1 mol </u> = x
30.97g )
Then you plug the answers into a calculator and you should get the answer of 0.74265.
You need to know the following equations:
pH=-log[H⁺]
pOH=-log[OH⁻]
pH+pOH=14
then you can find the pH by using pH=-log(1×10⁻¹)
pH=1
then you can find pOH by doing pOH=14-1
pOH=13
then you can find [OH] by using [OH⁻]=10^(-pOH)
[OH⁻]=1×10⁻¹³mol/L
You could also use the equation Kw=[OH⁻][H⁺] where Kw=10⁻¹⁴
10⁻¹⁴/10⁻¹=10⁻¹³mol/L
I hope this helps. Let me know if anything is unclear.
Answer:The weight of Object F will be times the weight of Object E if both objects are placed on the same planet.
Explanation:
Mass of the object E,= 4800 kg
Mass of the object F,== 600 kg
Since they are on the same planet same gravitation force will be exerted on them by the gravity of the planet.
The weight of Object F will be times the weight of Object E if both objects are placed on the same planet.
Answer:
pH = 8.92
Explanation:
To solve this question we must know that the reaction of KOH with HC3H5O2 is:
KOH + HC3H5O2 → H2O + KC3H5O2
At equivalence point, all propanoic acid reacts to produce KC3H5O2.
This KC3H5O2 = C3H5O2⁻ is in equilibrium in water as follows:
C3H5O2⁻(aq) + H₂O(l) → OH⁻(aq) + HC3H5O2(aq)
<em>Where Kb = Kw / ka = 1x10⁻¹⁴/ 1.34x10⁻⁵ = 7.46x10⁻¹⁰</em>
is defined as:
Kb = 7.46x10⁻¹⁰ = [OH⁻] [HC3H5O2] / [C3H5O2⁻]
As both [OH⁻] [HC3H5O2] ions comes from the same equilibrium,
[OH⁻] = [HC3H5O2] = X
[C3H5O2⁻] is:
<em>Moles KOH = Moles </em>C3H5O2⁻:
0.0325L * (0.15mol / L) = 0.004875 moles
In 32.5 + 20mL = 52.5mL = 0.0525L:
0.004875 moles / 0.0525L = 0.09286M.
Replacing:
7.46x10⁻¹⁰ = [X] [X] / [0.09286M]
6.927x10⁻¹¹ = X²
X = 8.323x10⁻⁶M = [OH-]
As pOH = -log [OH-]
pOH = 5.08
pH = 14 -pOH
<h3>pH = 8.92</h3>