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Ivenika [448]
3 years ago
7

Compare and contrast how observations and results can be used to

Chemistry
2 answers:
worty [1.4K]3 years ago
5 0

Answer:

by statistical analyses, especially by determining the p-value

Explanation:

In general, observations and results obtained from experimental procedures are subjected to a statistical test to check the robustness of the working hypothesis. The p-value is the most widely used statistical index in order to test such observations and results. The p-value is the statistical probability of obtaining extreme observed results when the null hypothesis is considered correct. A p-value lesser than 0.05 generally is considered statistically significant and then the null hypothesis can be rejected. In consequence, a very low p-value (which is obtained by statistical analysis of the observations and results), indicates that there is strong evidence in support of the alternative hypothesis.

dusya [7]3 years ago
4 0

ANSWER: by statistical analyses, especially by determining the p-value

EXPLANATION:                                                                                                                   In general, observations and results obtained from experimental procedures are subjected to a statistical test to check the robustness of the working hypothesis. The p-value is the most widely used statistical index in order to test such observations and results. The p-value is the statistical probability of obtaining extreme observed results when the null hypothesis is considered correct. A p-value lesser than 0.05 generally is considered statistically significant and then the null hypothesis can be rejected. In consequence, a very low p-value (which is obtained by statistical analysis of the observations and results), indicates that there is strong evidence in support of the alternative hypothesis.

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PLEASE HELP!!!!!!!!!
andreev551 [17]

The average atomic mass of the imaginary element : 47.255 amu

<h3>Further explanation  </h3>

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

Mass atom X = mass isotope 1 . % + mass isotope 2.% ..

isotope E-47 47.011 amu, 87.34%

isotope E-48 48.008 amu, 6.895

isotope E-49 50.009 amu, 5.77%

The average atomic mass :

\tt avg~mass=0.8734\times 47.011+0.06895\times 48.008+0.0577\times 50.009\\\\avg~mass=41.059+3.310+2.886\\\\avg~mass=47.255~amu

5 0
2 years ago
1. A contour line represents a single equal elevation: that is, all points on the same contour line have the same elevation.
Lady bird [3.3K]
The answer is hard to give without any information.
3 0
1 year ago
Consider a 0.10 M aqueous benzoic acid, CeHeCOOH. The K benzoic acid. 6.5 x 10 for A) Write a balanced equation that shows the r
7nadin3 [17]

Answer:

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

c) pH = 2.599

Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

7 0
3 years ago
Provide the element name with a ground-state electron configuration of 1s22s2p4
Reika [66]
Oxygen I’m pretty sure
8 0
3 years ago
The three states of matter differ primarily in terms of shape and volume. Describe solids, liquids and gases in these terms.
schepotkina [342]

Answer:

solids have definite shape and volume. liquid have definite volume but not definite shape. gases do not have definite volume as well as definite shape.

8 0
3 years ago
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