Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
Answer: 0.5 g/cm^3
Density equals mass divided by volume so..
60/120 is 0.5 g/cm^3
<span>Many scientific investigations have provided evidence to support this as the best explanation of the data</span>
Answer:
+1.03 V
Explanation:
The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).
The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:
Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V
Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V
The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.
emf = E°reduces - E°oxides
emf = 0.80 - (-0.23)
emf = +1.03 V
A glow stick will glow longer at lower temperatures than at room temperature, one can infer from the observation. Temperature and reaction time are the test variables.
We notice in this reaction that a glow stick stored in the freezer lights for a longer period of time than a glow stick stored at normal temperature. This implies that temperature affects how long a response lasts.
The most straightforward explanation for this observation is that glow sticks glow longer in colder temperatures than they do at room temperature; as a result, glow sticks kept in the freezer are observed to glow longer than glow sticks kept at room temperature.
To learn more about chemicals to the given link:
brainly.com/question/24600141
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