Answer: Y and Z. It's D.
Explanation: W represents evaporation, which is a physical change. X represents dissolving, which is also a physical change. If X and W aren't chemical changes, then Y and Z must be if half of the choices are.
Answer:
a) 0.03050 = 3.050 × 10-²
b) 0.256 x 10°= 2.56 × 10-¹
c) 25.005 10 = 2.500510 × 10¹
Explanation:
Scientific notations is a way of making very large or very small numbers more comprehensive or simplified. It involves the use of power of ten (10^). The numbers are represented to the power of ten. The following format is used:
a x 10^b
where; a is a number or decimal number between 1 and 10 i.e less than 10 but greater than 1
b is the power of ten
To write a number in scientific notation,
- we move the decimal point right or left depending on whether we're trying to reduce or increase the number
- we count the number of times the decimal point was moved. This serves as the b in the format above.
For example,
a) 0.03050 = 3.050 × 10-²
The decimal point was moved rightward twice. This caused the ^-2 power.
b) 0.256 x 10°= 2.56 × 10-¹
The decimal point was moved rightward once. This caused the ^-1 power.
c) 25.005 10 = 2.500510 × 10¹
The decimal point was moved leftward once. This caused the ^1 power.
Answer:
Explanation
Comment
Normally you would just use m*c*delta T. But the c is the same in both cases, so you need just use m*deltaT
Givens
25 degrees
m = ?
T = 25
t = final temp = 75
93 degrees
m = 250
T = 93
t = final temp = 75
Solution
(t - 25)*m = (93 - t) * 250 Remove the brackets
75*m - 25m = 93*250 - 250*75 Combine the right
m(75 - 25) = 23250 - 250*75
50 m = 23250 - 18750 Combine
50 m = 4500 Divide both sides by 50
50*m/50 = 4500/50
m = 90
Answer: 90 grams are needed.
Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V
