<span>Now consider a low pressure area on a disk as shown below.A parcel of air at point A would move toward the center of the low pressure area. That movement would take it farther away from the center of the disk and therefore it would move to the west. A parcel of air at B would move toward the center of the low pressure area which would also take it closer to the center of the spinning disk where its speed is greater than the surrounding points. It would appear to move to the east. With A moving to the west and B moving to the east the line from A to B is rotating counterclockwise.</span>
V
1
/T
1
=V
2
/T
2
(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.
Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
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)
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.