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Bess [88]
3 years ago
15

Why is it important for scientist to be repeatable?

Chemistry
2 answers:
shutvik [7]3 years ago
8 0

Answer:

It is important for scientists to be repeatable so it lets them see patterns and trends in their results. This is affirmative for their work, making it stronger and better able to support their claims. This helps maintain The integrity of data.

Repeating an experiment more than once helps determine if the data was a fluke, or represents the normal case. It helps guard against jumping to conclusions without enough evidence. The number of repeats depends on many factors, including the spread of the data and the availability of resources

Explanation:

Margarita [4]3 years ago
3 0
It is important for scientists to be repeatable so it lets them see patterns and trends in their results. This is affirmative for their work, making it stronger and better able to support their claims. This helps maintain integrity of data.
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Acid/base, precipitation, redox, or none:. 1. Ba(ClO3)2(s) ----> BaCl2(s) + 3O2(g). 2. CaO(s) +CO2(g) ----> CaCO3(s)
Aleks04 [339]
There are two equations given in the question. The type of reaction shown can be easily judged, if one takes a closer look at the equations.

1. <span>Ba(ClO3)2(s) ----> BaCl2(s) + 3O2(g) ------- Redox

2. </span><span>CaO(s) +CO2(g) ----> CaCO3(s) -------- None.

I hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>
4 0
3 years ago
The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av
nlexa [21]

<u>Answer:</u> The percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{77}^{191}\textrm{Ir} isotope be 'x'. So, fractional abundance of _{77}^{193}\textrm{Ir} isotope will be '1 - x'

  • <u>For _{77}^{191}\textrm{Ir} isotope:</u>

Mass of _{77}^{191}\textrm{Ir} isotope = 190.9606 amu

Fractional abundance of _{77}^{191}\textrm{Ir} isotope = x

  • <u>For _{77}^{193}\textrm{Ir} isotope:</u>

Mass of _{77}^{193}\textrm{Ir} isotope = 192.9629 amu

Fractional abundance of _{77}^{193}\textrm{Ir} isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710

Percentage abundance of _{77}^{191}\textrm{Ir} isotope = 0.3710\times 100=37.10\%

Percentage abundance of _{77}^{193}\textrm{Ir} isotope = (1-0.3710)=0.6290\times 100=62.90\%

Hence, the percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

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3 years ago
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