<u>Answer:</u>
<em> needed to heat the amount of water giving.
</em>
<u>Explanation:</u>
To solve this question we first need to know the specific heat of water which is given by <em>4.18 joules</em> Per gram for increasing <em>1° of temperature</em>. Now since one gram of water name one joules per gram so 2 grams will need <em>twice of 4.18</em> that is 8.36 similarly 50 grams will need
Now we need to calculate the heat for 10° of change because temperature rises from 50° centigrade to 60° centigrade and difference in degrees
<em>so heat needed is= </em>
The balanced equation for the redox reaction, which takes place in a basic solution is:
- 2 Cr(OH)₃ + ClO₃⁻ (aq) + 4 OH⁻ → 2 CrO₄²⁻ (aq) + Cl⁻ (aq) + 5 H₂O; <u>option C</u>
<h3>What is the balanced equation for the redox reaction?</h3>
A balanced equation for a redox reaction is one in which the electrons transferred are balanced and the atoms involved in the reaction are also balanced.
To balance redox reactions in a basic solution, OH⁻ ions and H₂O are placed on appropriate sides of the reaction.
Considering the given reaction:
Cr(OH)₃ + ClO₃⁻ (aq) → CrO₄²⁻ (aq) + Cl⁻ (aq):
The balanced equation will be:
2 Cr(OH)₃ + ClO₃⁻ (aq) + 4 OH⁻ → 2 CrO₄²⁻ (aq) + Cl⁻ (aq) + 5 H₂O
The electrons transferred and the atoms involved are balanced.
Learn more about redox reactions at: brainly.com/question/21851295
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Answer:-
Carbon cannot reduce sodium, magnesium and calcium to their respective metals because carbon is lower than them in the electrochemical series.
Only metals higher in the electrochemical series can displace metals placed lower.
Sodium, magnesium and calcium are obtained from their salts by electrolysis of the salts.
Sodium Chloride is the sodium salt used. Sodium is deposited at cathode.
Chlorine gas is produced from the process.
Cathode: 2Na+ + 2e- --> 2Na (Reduction)
Anode: 2Cl- --> Cl2 + 2e- (Oxidation)
Potassium and Calcium are two other metals that can be extracted by the same method
From the molarity and volume of HClO4, we can determine how many moles of H+ we initially have:
0.18 M HClO4 * 0.100 L HClO4 = 0.018 moles H+
We can determine how many moles of OH- we have from the molarity and volume of LiOH:
0.27 M LiOH * 0.030 L LiOH = 0.0081 moles OH-
When the HClO4 and LiOH neutralize each other, the remaining will be
0.018 moles H+ - 0.0081 moles OH- = 0.0099 moles of excess H+
This means that the molarity [H+] will be
[H+] = 0.0099 moles H+ / (0.100 L + 0.030 L) = 0.07615 M
The pH of the solution will therefore be
pH = -log [H+] = -log 0.07615 = 1.12
They have similar electron configurations and have similar ionic chargers