To determine the number of cups of milk, we first calculate for the volume of the milk needed. Then, we use a conversion factor for the volume from cubic centimeter to cups. From literature, 1 cubic centimeter is equal to 0.0042 cup. We do as follows:
Volume of milk = ( 2.50 kg ) ( 1000 g / 1 kg ) / 1.03 g /cm^3 = 2427.18 cm^3
cups of milk = 2427.18 cm^3 ( 0.0042 cup / 1 cm^3 ) = 10.19 cups
Answer:
In addition to mathematics, physics and astronomy, Newton also had an interest in alchemy, mysticism, and theology.
The chief advantage of the metric system over other systems of measurement is that it B. is in multiples of 10.
This can be seen in the picture below that shows the prefixes of the metric system.
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
