Answer:
This question is incomplete
Explanation:
The question is incomplete because of the absence of the table but since the question says there are data from an investigation about a plant growth and five other plants (making six) of the same type, the best way to display this type of data for analyst is to use the grouped bar chart. <u>The grouped bar chart will display the data obtained (from an investigation on plant growth) from different students on each of the six plants (of the the same type)</u>.
Colours are usually used to identify the bars (of a group) or could be used to separate the group from other groups but in this case, colours are better used to identify the bars of a group.
Answer:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Answer:
3Mg(NO3)2(aq)+2Na3PO4(aq)⇒Mg3(PO4)2(s)+6NaNO3(aq)
Explanation:
Answer:
Option D. 30 mL.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above,
The mole ratio of the acid, nA = 1
The mole ratio of the base, nB = 1
Step 2:
Data obtained from the question. This include the following:
Volume of base, KOH (Vb) =.?
Molarity of base, KOH (Mb) = 0.5M
Volume of acid, HNO3 (Va) = 10mL
Molarity of acid, HNO3 (Ma) = 1.5M
Step 3:
Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:
MaVa / MbVb = nA/nB
1.5 x 10 / 0.5 x Vb = 1
Cross multiply
0.5 x Vb = 1.5 x 10
Divide both side by 0.5
Vb = (1.5 x 10) /0.5
Vb = 30mL
Therefore, the volume of the base, KOH needed for the reaction is 30mL.
