The answer is , 1- a 2-c 3- a 4- a 5- d
Complete question :
A jeweler has five rings, each weighing 18 g, made of an alloy of 10% silver and 90% gold. She decides to melt down the rings and add enough silver to reduce the gold content to 75%. How much silver should she add?
Answer:
Kindly check explanation
Explanation:
Given the following :
Weight of each alloy = 18g
Initial % of Silver = 10% = 0.1
Initial % of Gold = 90% = 0.9
New % of gold = 75% = 0.75
Hence, new % of silver will be (1 - 0.75) = 0.25
Let the amount of silver to be added = s
Hence,
Initial gold content = final gold content + added silver
18 * 0.9 = 0.75(18 + s)
16.2 = 13.5 + 0.75s
16. 2 = 13.5 + 0.75s
16.2 - 13.5 = 0.75s
2.7 = 0.75s
s = 2.7 / 0.75
s = 3.6
S = 3.6
Hence the amount of silver needed per ring in other to reduce the gold content = 3.6g
Total for the 5 rings = (3.6 * 5) = 18g
Since 20 C is 68 F it would be safe to say that its at room temp. since your are just miking them there chemical properties stay the same and the iron would remain magnetic Hope this helps :)
Answer:
The answer to your question is: letter B
Explanation:
The information given is showing us three kinds of carbons and their respective number of protons and neutrons. Now, let's analize the options given:
A.)Carbon Bond Types This option is about trhe kind of bonds carbon is able to form like single, double or triple and is different from the information descrive above. So ,this option is incorrect.
B.)Carbon Isotopes is option is about the carbon atoms that have the same number of protons and different amount of neutrons exactly as it is describe above, to this option is correct.
C.)The Carbon Cycle
this option is about how carbon goes from one kind of matter to another and is completely different to the description given above, so this option is incorrect.
D.)Chemical Reactions this option is about reactants that after a process give products they could be from carbon -12, carbon-13 or carbon-14 and other elements, so this option is different from the description given above.
