Answer:
Chemicals can attach more easily to soils with a higher content of clay and organic matter. In addition, sandy soils generally do not contain a large amount of soil organisms compared with other soil types.
Explanation:
Make a quick chart with each element represented, and count them up. HINT - leave the polyatomic anions together - in this case, PO4
Left Right
1 Ca 3
2 O 1
5 H 2
1 PO4 2
Begin by balancing like finding common denominators of fractions - apply to both sides:
I started by adding a 2 in front of H3PO4 on the left, them 6 in front of H2O on the right. Last, a 3 in front of Ca (OH)2. Then, re-count using the chart format to make sure you're right.
3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O
Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
1. Decreases by 4. (B)
2. The atomic number changes. (B)
3. 56/26 Fe. (C)
4. Potassium-40;t1/2=25 days. (B)
5. Takes place in the upper atmosphere. (A)
Forces are pushes and pulls that may change the motion of an object. Balanced forces result in an object remaining at rest or moving at a constant speed. Unbalanced forces result in the acceleration of an object. An object's motion depends on how it changes position.
