Answer: 3.0 kJ × 1 mol/40.65 kJ× 18.02 g/mol × 1 mL/1 g= 1.3 mL
First let's convert the minutes to hours (multiply minutes by 60 to get hours):
30 × 60 = 1,800 drops/hour
Now that you know how much drops there are per hour, you can multiply this answer by 3 to work out how many drops there are in 3 hours:
1,800 × 3 = 5,400 drops
We know that 5 drops is equal to 1 ml, so we can divide 5,400 by 5 to work out the amount of ml:
5,400 ÷ 5 = 1,080 ml
Therefore, your final answer is 1,080 millilitres (ml)
Q=mc(deltaT)
Q is the amount of energy which you are looking for
M is the mass which you can find
C is the specific heat of water which is 4.18 J/gC
DeltaT is the change in temperature which you can find.
To find the mass, first you must know that the density of water is 1g/mL, meaning that 200 mL has a mass of 200 g. This means that to find the total mass (m in the equation) all you need to do is add the mass of water and NaOH.
200 g + 2.535 g=202.535 g.
To find deltaT you would need to take the final temperature minus the initial temperature.
27.8C-24.2C=3.6C
Then these values can be substituted into the equation:
q=(202.635g)(4.18J/gC)(3.6C)
Q=3049.25 J
Technically this should be rounded off to 1 significant figure (200 mL only had 1), but ignoring signficiant figure rules this should be correct. Also, sometimes other units like calories or kJ may be asked for, meaning that a conversion or alternate c value would be used.
Hey there!:
HA <=> H⁺ + A⁻
pH = -log[H+] = 6
[ H⁺ ] = 10^-pH
[ H⁺] = 10 ^ -6
[ H⁺ ] = 0.000001 M
Percent dissociation:
[ H⁺ ] / [ HA]o * 100
[ 0.000001 / 0.10 ] * 100
0.00001 * 100 => 0.0010%
Answer D
Hope that helps!
Answer:
Oxidation: a type of chemical reaction where one or more electrons are lost.
Oxidation State / Number: a number assigned to an atom describing its degree of oxidation, meaning how many electrons it has gained or lost.
Reduction: a type of chemical reaction where one or more electrons are gained.
Oxidation-Reduction Reaction: a chemical reaction where oxidation and reduction occurs simultaneously
Explanation:
Reduction always occurs at cathode
Oxidation always occurs in anode
These two process occurs in same way independent of nature of cell whether voltaic or electrolytic.
