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Vladimir [108]
3 years ago
12

The table shows data for two groups of plants one grown with fertilizer and the other without fetalizer was the mean height of t

he plant given fertilizer​
Chemistry
2 answers:
Yuki888 [10]3 years ago
7 0
Do you need variables orrr what is the question ?
AleksAgata [21]3 years ago
6 0

Answer:

What is the question?

Explanation:

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Use the balanced equation below
elixir [45]
You know oxygen is the limiting reactant (since it says there is excess hydrogen). So, use stoichiometry based on the given number of oxygen moles:

4 0
3 years ago
What is the kb of the base po−34 given that a 0. 48m solution of the base has a ph of 12. 70? the equation described by the kb v
podryga [215]

The equation described by the kb value is 5.21 x 10^-^3.

The potential of Hydrogen is what pH is formally known as. The negative logarithm of the concentration of H+ ions is known as pH. Thus, the definition of pH as the amount of hydrogen is provided. The hydrogen ion concentration in a solution is described by the pH scale, which also serves as a gauge for the solution's acidity or basicity.

Assuming  PO₄³⁻ (the phosphate anion).

PO₄³- + H₂O ==> HPO₄²⁻ + OH⁻

Kb = [HPO₄²⁻][OH⁻] / [PO₄³⁻]

We can find the [OH⁻] from the pH of 12.70.

pH + pOH = 14

14.0 - 12.7 = 1.3 = pOH

[OH] = 1x10^-^1^.^3

[OH-] = 5.0x10^-^2 M

[HPO₄²⁻] = 5.0x10^-^2 M

Kb = (5.0x10^-^2)2 / 0.48

     = 2.5x10^-^3 / 0.48

Kb = 5.21 x 10^-^3

Learn more about pH here:

brainly.com/question/8758541

#SPJ4

3 0
1 year ago
How would you prepare 1 liter of .030 m copper 2 sulfate pentahydrate?
Ierofanga [76]
Answer is: dissolve 74,9 grams CuSO₄·5H₂O in one liter volumetric flask.
V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O)  = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
n(CuSO₄·5H₂O) = 1 L · 0,3 mol/L.
n(CuSO₄·5H₂O) = 0,3 mol.
m(CuSO₄·5H₂O) = n(CuSO₄·5H₂O) · M(CuSO₄·5H₂O).
m(CuSO₄·5H₂O) = 0,3 mol · 249,7 g/mol.
m(CuSO₄·5H₂O) = 74,9 g.
5 0
3 years ago
Give the name of the element that is a member of the alkali metal family whose most stable ion contains 2 electrons.
notka56 [123]
Lithium

Li → Li⁺ + e⁻

Li 1s²2s¹ → Li⁺ 1s² + e⁻
3 0
4 years ago
El número de oxidación del N en el NaNO3 es:
horrorfan [7]

Answer:

Nose

Explanation:

7 0
3 years ago
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