Answer:
<h3>F=4k.gm/s^2</h3>
Explanation:
<h3>F=m×a</h3><h3>f=2k.g×2m/s^2</h3><h3>f=4k.gm/s^2</h3>
Answer: The pH of solution is 10.
The pOH of the solution is 4.
Explanation:
pH is the negative logarithm of concentration of hydrogen ion.
As given concentration of acidic solution is 
. Therefore, pH of the solution is calculated as follows.
![pH = -log [H^{+}]\\= -log (1.00 \times 10^{-10})\\= 10](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C%3D%20-log%20%281.00%20%5Ctimes%2010%5E%7B-10%7D%29%5C%5C%3D%2010)
The relation between pH and pOH is as follows.
pH + pOH = 14
pOH = 14 - pH
= 14 - 10
= 4
Thus, we an conclude that pH of solution is 10 and pOH of the solution is 4.
Answer:
A
Explanation:
because if you bond postitive and positive they would repell
As given:
Initial moles of P taken = 2 mol
the products are R and Q
at equilibrium the moles of
R = x
total moles = 2 + x/2
Let us check for each reaction
A) P <-> 2Q+R
Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = 2x
moles of R = x
Total moles = (2-x) + 2x + x = 2 +2x
B) 2P <-> 2Q+R
Here x moles of P will give x moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x
moles of R = x/2
Total moles = (2-x) + x + x/2 = 2 + x/2
C) 2P <-> Q+R
Here x moles of P will give x/2 moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x /2
moles of R = x/2
Total moles = (2-x) + x + x = 2
D) 2P <-> Q+2R
Here x moles of P will give x/2 moles of Q and x moles of R
So at equilibrium
moles of P left = 2-x
moles of Q = x/2
moles of R = x
Total moles = (2-x) + x/2 + x = 2 + x/2
Answer:
Anything that cannot be separated physically or chemically is an element.
