D has a total of four significant figures.
Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).
Answer:
Moles of KOH in 1000 mL solution = 0.255 moles
Moles of KOH in 1 mL solution = 0.255/1000 = 0.000255 moles
Moles in 95 mL solution = (95 * 255)/1000000 = 24225/1000000
Moles of KOH in 95 mL 0.255M solution = 0.024225 moles
Explanation:
The given data is as follows.
Energy of radiation absorbed by the electron in hydrogen atom = 
As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.
E = 
=
= 
or,
=
It is known that, 
= 
And, according to De-Broglie equation 
as, p = 
So, 
= 
Now, on squaring both the sides we get the following.
=
=

where, m = mass of electron
So, 
= 
=
J
Since, K.E = 
= 
= 
Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is
.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)