Answer:
64,433.6 Joules
Explanation:
<u>We are given</u>;
- Volume of water as 220 mL
- Initial temperature as 30°C
- Final temperature as 100°C
- Specific heat capacity of water as 4.184 J/g°C
We are required to calculate the amount of heat required to raise the temperature.
- We know that amount of heat is calculated by;
Q = mcΔT , where m is the mass, c is the specific heat, ΔT is the change in temperature.
Density of water is 1 g/mL
Thus, mass of water is 220 g
ΔT = 100°C - 30°C
= 70°C
Therefore;
Amount of heat, Q = 220g × 4.184 J/g°C × 70°C
= 64,433.6 Joules
Thus, the amount of heat required to raise the temperature of water is 64,433.6 Joules
Answer:
The answer to the question is
The height of the mercury fluid column remain the same.
Explanation:
The pressure, P in a column of fluid of height, h is given by
P = (Density p)×(height of fluid column h)×(gravity g)
Therefore, when the diameter is doubled we have
Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence
P = p×h1×g = p×h2×g Therefore h1 = h2
The height of the fluid column remain the same
Answer:
the answer is letter B,FALSE
<span>I'm pretty sure it because of chemical change. :) Hope this helps!</span>
Reduction takes place in cathode while oxidation takes place at the anode.
Given that the reaction 3MnO4- +24H + +5Fe→3Mn+2+5Fe+3+12H2O
Now that oxidation is termed as an increase in oxidation number or loss of electrons while reduction is a decrease in oxidation number and gain of electrons,
∴oxidation will be
Fe→Fe +3+3e-
reduction will
MnO4- +8H+ →Mn+2+4H2O
If oxidation takes place at anode then Anode: an oxidation reaction
Fe→Fe+3+3e- . Then the answer is
Fe(s)→Fe+3(aq)+3e-
