It involves breaking molecular bonds between copper compounds.
<span>From Beer-Lambert's law' A = ECl Since the slope is 1550.7
thus A = 1550.7* 0.000529 M = 0.8203203 Since A = Log (Io/I) and T% = I/Io * 100% Io/I = AntiLog(A) and hence T% = { 1/ AntiLog (A) } *100%
T%={1/AntiLog 0.8203203)*100% T%=0.8601654108*100% T%= .860</span>
Electrolysis of water involves the decomposition of water into oxygen and hydrogen gas by passing an electric current through water. The positively charged ion (hydrogen ion) moves to the negative electrode (cathode) during electrolysis, while the negatively charged ion (OH-) moves to the positive electrode (anode). Therefore, the half equation that occurs at the cathode during the electrolysis of water will be;
4H+(aq) + 4e- = 2H2 (g)
Answer:
If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial
In this case:
- Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)
- m=700 g
- ΔT= Tfinal - Tinitial= Tfinal - 90 °C
Replacing:
Solving:
16.125 °C= Tfinal - 90 °C
Tfinal= 16.125 °C + 90 °C
Tfinal= 106.125 °C
<u><em>If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C</em></u>
Correct question is;
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is?
Answer:
6%
Explanation:
From ohms law, we know that;
R = V/I
Where;
R is resistance
V is voltage
I is current
Now, the percentage error in the resistance is given by the formula;
ΔR/R = ΔV/V+ ΔI/I
We are told that the current and the voltage difference have a percentage error of 3% each. Thus;
ΔR/R = 3% + 3% = 6%