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Law Incorporation [45]
3 years ago
8

What are the differences between sand and potting soil? Are they both mixtures? How do you know?

Chemistry
1 answer:
katen-ka-za [31]3 years ago
7 0
What class is this for because it depends
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Volcano experiment
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3 years ago
A(n)<br> wave carries energy through matter.
denis23 [38]

Answer:

A wave is a disturbance that carries energy from one place to another through matter and space. When we through a stone or a pebble in calm water, then the particles of water moves up and down and this process continues for some time. This implies that there is a disturbance produced in water.

Explanation:

4 0
3 years ago
The average atomic mass recorded on the periodic table for cobalt is 58.933 u. This indicates that the most abundant isotope of
OleMash [197]
The average atomic mass of the element is the sum of the products of the percentage abundance of isotope and its mass number. Therefore, for atomic mass equal to 58.933, the most abundant isotope is cobalt-59. Thus, the answer is letter C. 
7 0
3 years ago
The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st
Andrei [34K]
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2  so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄      ⇄     H⁺   + HC₂O₄⁻
0.0356 M            0          0
0.0356 - x            x          x
Ka1 = \frac{[H^+][HC2O4^-]}{[H2C2O4]} = x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
5 0
2 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
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