Answer:
C. 0.001 second
Explanation:
Millisecond is a subunit of a second.
Milli depicts a unit to the power of 10⁻³
So;
1 millisecond = 10⁻³ s
Therefore,
1 x 10⁻³ = 0.001s
The solution to this problem is choice C which is 0.001 second
Answer:
3.99 g
Explanation:
The following data were obtained from the question:
Half life (t½) = 2 years.
Original amount (N₀) = 128 g
Time (t) = 10 years
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 years.
Decay constant (K) =.?
K= 0.693/t½
K = 0.693/2
K = 0.3465 year¯¹.
Finally, we shall determine the amount remaining after 10 years i.e the amount remaining when they arrive on Earth. This can be obtained as follow:
Original amount (N₀) = 128 g
Time (t) = 10 years
Decay constant (K) = 0.3465 year¯¹.
Amount remaining (N) =..?
Log (N₀/N) = kt/2.3
Log (128/N) = (0.3465 × 10)/2.3
Log (128/N) = 1.5065
Take the antilog of 1.5065
128/N = Antilog (1.5065)
128/N = 32.1
Cross multiply
128 = 32.1 × N
Divide both side by 32.1
N = 128/32.1
N = 3.99 g
Therefore, the amount remaining is 3.99 g
Carbon dioxide, water, energy
Answer:
A) Precision
B) Accuracy
C) Precision
D) Accuracy
Explanation:
<em>Accuracy</em> and <em>Precision </em>are factors that determine whether a given analysis is appropiate or not.
Accuracy refers to how close the experimental result is to a <em>known, theoretical or accepted value. </em>For this reason every method that checks the accuracy uses a standard -this could mean repeating the test using the standard instead of the sample, or adding a known amount of analyte to the sample-, in order to compare the results.
Precision refers to how close repeated the results of different tests are to each other, in other words, it refers to the <em>repeatability </em>of the method. A test for precision would require that the method is done several times, in order to check how close the results are to each other, regardless of whether those results are close to the expected value.
