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worty [1.4K]
3 years ago
15

Robots have been around since the 3rd century BC. At that time, they were mainly invented as toys to amuse the king or emperor.

Chemistry
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

Answer:

B. I and II only  

Explanation:

For example, unmanned space probes can gather information more easily than a human can.

Closer to home, many biochemical research and pharmaceutical industries use pipetting robots for their repetitive pipetting chores.

C is wrong.  Machines can do things faster than any human ever could and not get tired, but humans possess the intelligence, expertise, and skills to use the robots most efficiently and to solve problems when things go wrong.

timama [110]3 years ago
6 0
A I think sorry if it’s wrong
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The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later
Kipish [7]

Answer:

12.02 g

Explanation:

From the question given above, the following data were obtained:

Half life (t½) = 2 days

Original amount (N₀) = 96 g

Time (t) = 6 days

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 2

K = 0.3465 /day

Therefore, the rate of disintegration of the isotope is 0.3465 /day.

Finally, we shall determine the amount of the isotope remaining after 6 days as follow:

Original amount (N₀) = 96 g

Time (t) = 6 days

Decay constant (K) = 0.3465 /day.

Amount remaining (N) =.?

Log (N₀/N) = kt / 2.303

Log (96/N) = (0.3465 × 6) / 2.303

Log (96/N) = 2.079/2.303

Log (96/N) = 0.9027

Take the anti log of 0.9027

96/N = anti log (0.9027)

96/N = 7.99

Cross multiply

96 = N × 7.99

Divide both side by 7.99

N = 96 /7.99

N = 12.02 g

Therefore, the amount of the isotope remaining after 6 days is 12.02 g

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3 years ago
Which of the following is NOT an example of proxy data?
Tomtit [17]

Answer:

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Explanation:

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valkas [14]

Answer:

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Explanation:

3 0
3 years ago
Read 2 more answers
Balance the following reaction in acidic solution:. . Ag(s) + NO3-(aq) -> Ag+(aq) + NO(g)
nekit [7.7K]
The two half-reactions are...
Ag→Ag+
and...
NO3→NO
Let's start by balancing the first half-reaction...
Ag→Ag+
The amounts are already balanced; 1:1. The oxygens are balanced. So all that's left is to balance the charge...
Ag→Ag++e−
Now let's do the other equation... Amounts of nitrogen are balanced, so we first need to balance the oxygens...
NO3→NO
4H++NO3→NO+2H2O
Next, we need to balance charge...
4e−+4H++NO3→NO+2H2O
Now let's go ahead and rewrite each half-reaction after being balanced by themselves...
Ag→Ag++e−
4e−+4H++NO3→NO+2H2O
Now we need to multiply by some factor to get the electrons to cancel out. In this case, that factor is 4, which needs to be applied to the top half-reaction...
4(Ag→Ag++e−)=4Ag→4Ag++4e−
Then we combine this half-reaction with the second one above to get...
4Ag+4H++NO3→4Ag++NO+2H2O
3 0
2 years ago
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