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skelet666 [1.2K]
2 years ago
5

The reliability or reproducibility of a measurement is its ...

Chemistry
1 answer:
Inessa [10]2 years ago
3 0

Answer:

The two most important aspects of precision are reliability and validity. Reliability refers to the reproducibility of a measurement. You quantify reliability simply by taking several measurements on the same subjects!

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In your video respond to the following question/promt. Make sure you state a clear claim and explain with
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Answer:

Vvvbn n njjnjj

Explanation:

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This is all no chemistry but the answer is C
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Answer these questions based on 234. 04360 as the atomic mass of thorium-234. The masses for the subatomic particles are given.
nikdorinn [45]

The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.

<h3>How do we calculate atomic mass?</h3>

Atomic mass (A) of any atom will be calculated as:

A = mass of protons + mass of neutrons

In the Thorium-234:

Number of protons = 90

Number of neutrons = 144

Mass of one proton = 1.00728 amu

Mass of one neutron = 1.00866 amu

Mass of thorium-234 = 90(1.00728) + 144(1.00866)

Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu

Given mass of thorium-234 = 234.04360 amu

Mass defect = 235.90224 - 234.04360 = 1.85864 amu

Hence required value is 1.85864 amu.

To know more about Atomic mass (A), visit the below link:

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4 0
1 year ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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