Answer:
answer willl be gold
Explanation:
bcz 1st highest thermal conductivity is having solver
then it is followed by copper , gold
so gold will be the answer
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
Answer:
0.170 M
Explanation:
M1V1=M2V2
(25.0mL)(0.123M)=(18.04mL)(x)
x= 0.170 M
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Answer: Lithium
Explanation: The balanced chemical equation is:
It can be seen, 4 moles of lithium combines with 1 mole of oxygen gas to produce 2 moles of lithium oxide.
Thus 8.4 moles of lithium combines with=of oxygen gas to produce 4.2 moles of lithium oxide.
As, Lithium limits the formation of product, it is the limiting reagent and Oxygen gas is present in excess, it is called the excess reagent. (4.6-2.1)=2.5 moles of oxygen gas are present in excess.
First, by magnet separate paper clips.
Then, By handpicking, separate the pebbles.
Then, using a suitable sieve, separate toothpicks and toothpicks.