<span>Friction creates heat which in turn can lead to deviations from the original size and shape of a part.
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Kilometers, Meters and centimeters if metric
Feet, inches, yards and miles if customary ( u.s.)
Answer:
In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab
Explanation:
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.
(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.
The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.
(2) C(s) + O2(g) --> CO2(g)
The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.
The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.
(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺
The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.
The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.
Answer:
, zirconium-103.
Explanation:
In a nuclear reaction, both the mass number and atomic number will conserve.
Let 
represent the unknown particle.
The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle , 
is the mass number while 
while is the atomic number.
Sum of mass numbers on the left-hand side of the equation:
.
Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:
.
Mass number conserves. As a result,
.
Solve this equation for :
.
Among the five choices, the only particle with a mass number of 103 is . Make sure that atomic number also conserves.
