<span>Mutation. Either exchanging a Purine with another Purine, Pyrimidin with another Pyrimidin, or completely exchanging a Purine with a Pyrimidin or vice versa. Point- or Frameshift-Mutation.</span>
It would be a solid because at -20 °C it has turned to ice
Answer:
0.62 N
Explanation:
first of all: 3.0 s = 0.05 m
Let ? be the work that brakes do at 3.0s
24 m ----> 299 N
0.05 m ---->?
? = (0.05 m x 299 N)/ 24
? = (14.95)/ 24
? = 0.622 N
Answer:
- Option A) <u><em>Mg + Cl₂ → MgCl₂</em></u>
Explanation:
The law of conservation of mass is guaranteed in a chemical equation. Since the mass of the atoms do not change, that means that the number of each kind of atoms in the reactant side is equal to the number of atoms of the same kind in the product side.
The first equation is:
<em><u>A) Mg + Cl₂ → MgCl₂</u></em>
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Number of atoms:
atom Reactant side Product side
Mg 1 1
Cl 2 2
Therefore, the table displays that there are the same number of atoms of each kind on both sides, showing that<em> the total mass during the chemical reaction stays the same.</em>
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<em><u>B) NaOH + MgCl₂ → NaCl + MgOH</u></em>
This equation displays 2 atoms of Cl on the left side and 1 atom of Cl on the right side; thus, it is not showing that the total mass stays the same during the chemical reaction.
<em />
<u><em>C) 2Na + 2H₂O → NaOH + H₂</em></u>
Neither the sodium, nor oxygen, nor hydrogen atoms are balanced. Thus, this does not show that the total mass stays the same.
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<u><em>D) H₂O + O₂ → H₂O</em></u>
The reactant side contains 3 oxygen atoms and the product side contains 1 atoms of oxygen; thus, this is not balanced: it does not show that the total mass stays de same during the chemical reaction.
Answer:
1/32 of the original sample
Explanation:
We have to use the formula
N/No = (1/2)^t/t1/2
N= amount of radioactive sample left after n number of half lives
No= original amount of radioactive sample present
t= time taken for the amount of radioactive samples to reduce to N
t1/2= half-life of the radioactive sample
We have been told that t= five half lives. This implies that t= 5(t1/2)
N/No = (1/2)^5(t1/2)/t1/2
Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;
N/No= (1/2)^5
N/No = 1/32
Hence the fraction left is 1/32 of the original sample.
