The correct answer is option A. Energy cannot be created during an ordinary chemical reaction. There is no such thing as an ordinary chemical reaction. Energy cannot be created or destroyed this is according to the law of conservation of energy. It can only be transformed from one form to another form.
Answer:
molality of sodium ions is 1.473 m
Explanation:
Molarity is moles of solute per litre of solution
Molality is moles of solute per kg of solvent.
The volume of solution = 1 L
The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams
The mass of solute = moles X molar mass of sodium phosphate = 0.65X164
mass of solute = 106.6 grams
the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg
the molality = 
Thus molality of sodium phosphate is 0.491 m
Each sodium phosphate of molecule will give three sodium ions.
Thus molality of sodium ions = 3 X 0.491 = 1.473 m
Answer:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring = 8 g/mL
Explanation:
From the question given above, the following data were obtained:
Mass of ring = 32 g
Volume of water = 64 mL
Volume of water + ring = 68 mL
Density of ring =?
Next, we shall determine the volume of the ring. This can be obtained as follow:
Volume of water = 64 mL
Volume of water + ring = 68 mL
Volume of ring =?
Volume of ring= (Volume of water + ring) – (Volume of water)
Volume of ring = 68 – 64
Volume of ring = 4 mL
Finally, we shall determine the density of the ring. This can be obtained as follow:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring =?
Density = mass / volume
Density of ring = 32 / 4
Density of ring = 8 g/mL
Answer:
Part a: <em>Units of k is </em>
<em> where reaction is first order in A and second order in B</em>
Part b: <em>Units of k is </em>
<em> where reaction is first order in A and second order overall.</em>
Part c: <em>Units of k is </em><em> where reaction is independent of the concentration of A and second order overall.</em>
Part d: <em>Units of k is </em>
<em> where reaction reaction is second order in both A and B.</em>
Explanation:
As the reaction is given as
where as the rate is given as
![r=k[A]^x[B]^y](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5Ex%5BB%5D%5Ey)
where x is the order wrt A and y is the order wrt B.
Part a:
x=1 and y=2 now the reaction rate equation is given as
![r=k[A]^1[B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E2)
Now the units are given as
![r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E1%5BM%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B1%2B2%7D%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B3%7D%5C%5CM%5E%7B1-3%7D%2Fs%20%3Dk%5C%5CM%5E%7B-2%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is
Part b:
x=1 and o=2
x+y=o
1+y=2
y=2-1
y=1
Now the reaction rate equation is given as
![r=k[A]^1[B]^1](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E1)
Now the units are given as
![r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E1%5BB%5D%5E1%5C%5CM%2Fs%20%3Dk%5BM%5D%5E1%5BM%5D%5E1%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B1%2B1%7D%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B2%7D%5C%5CM%5E%7B1-2%7D%2Fs%20%3Dk%5C%5CM%5E%7B-1%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is
Part c:
x=0 and o=2
x+y=o
0+y=2
y=2
y=2
Now the reaction rate equation is given as
![r=k[A]^0[B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E0%5BB%5D%5E2)
Now the units are given as
![r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BB%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B2%7D%5C%5CM%5E%7B1-2%7D%2Fs%20%3Dk%5C%5CM%5E%7B-1%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is
Part d:
x=2 and y=2
Now the reaction rate equation is given as
![r=k[A]^2[B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E2%5BB%5D%5E2)
Now the units are given as
![r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E2%5BB%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E2%5BM%5D%5E2%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B2%2B2%7D%5C%5CM%2Fs%20%3Dk%5BM%5D%5E%7B4%7D%5C%5CM%5E%7B1-4%7D%2Fs%20%3Dk%5C%5CM%5E%7B-3%7Ds%5E%7B-1%7D%20%3Dk)
The units of k is
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.
