Your answer is A. Weathered rocks and decomposed organic remains.
V = 3.00 x 3.00 x 3.00 => 27 cm³
D = m / V
8.90 = m / 27
m = 8.90 x 27
m = 240.3 g
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer:None of the tunes make a correct statement the third choice is closed but may still dating the pencil operates broken because the light been to a zoo from a straight line when it across the boundary between air and water
Explanation:
PH of a solution is -ln[H3O+]
so,in case of A pH=3 or,-log[H3O+]=3 or,[H3O+]=10^-3
in case of B pH=6 pr,-log[H3O+]=6 or, [H3O+]=10^-6
so,hydronium ion concentration in solution A /the hydronium ion concentration in solution Z
=10^-3/10^-6
=1000
2)
Ca(OH)2+2 HNO3=Ca(NO3)2+2 H2O
so the answer is 2.
