The variables to control in an experiment to determine the effect of different fertilizers on the rate of plant growth include soil composition, temperature, water, and light.
<h3>What are controlled variables?</h3>
The expression controlled variables makes reference to experimental conditions that must be equal or constant between experimental groups in order to obtain better comparisons when collecting results.
In conclusion, The variables to control in an experiment to determine the effect of different fertilizers on the rate of plant growth include soil composition, temperature, water, and light.
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Answer:
The correct answer is A) single
Explanation:
In the case of the Bromo atom, it requires 1 electron to complete its octet, therefore it shares 1 electron with the other Bromo atom.
<u>Answer:</u> The chemical equations are given below.
<u>Explanation:</u>
The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:
By Stoichiometry of the reaction:
1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.
The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:
![Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^{2-}(aq.)](https://tex.z-dn.net/?f=Pb%28OH%29_2%28s%29%2B2OH%5E-%28aq.%29%5Crightarrow%20%5BPb%28OH%29_4%5D%5E%7B2-%7D%28aq.%29)
By Stoichiometry of the reaction:
1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex
Hence, the chemical equations are given above.
Answer:
a) pH = 13.176
b) pH = 13
c) pH = 12.574
d) pH = 7.0
e) pH = 1.46
f) pH = 1.21
Explanation:
HBr + NaOH ↔ NaBr + H2O
∴ equivalent point:
⇒ mol acid = mol base
⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)
⇒ Va = 0.025 L
a) before addition acid:
⇒ <em>C </em>NaOH = 0.150 M
⇒ [ OH- ] = 0.150 M
⇒ pOH = - Log ( 0.150 )
⇒ pOH = 0.824
⇒ pH = 14 - pOH
⇒ pH = 13.176
b) after addition 5mL HBr:
⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M
⇒ <em>C </em>HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M
⇒ [ OH- ] = 0.1 M
⇒ pOH = 1
⇒ pH = 13
c) after addition 15mL HBr:
⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M
⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M
⇒ [ OH- ] = 0.0375 M
⇒ pOH = 1.426
⇒ pH = 12.574
d) after addition 25mL HBr:
equivalent point:
⇒ [ OH- ] = [ H3O+ ]
⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²
⇒ [ H3O+ ] = 1 E-7
⇒ pH = 7.0
d) after addition 40mL HBr:
⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M
⇒ [ H3O+ ] = 0.035 M
⇒ pH = 1.46
d) after addition 60mL HBr:
⇒ <em>C</em> HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M
⇒ [ H3O+ ] = 0.062 M
⇒ pH = 1.21
Answer:
26,000 KL/kg will be the density of the Elephant.
