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4 years ago

During which part of the scientific method would error bars be used?

Chemistry

1 answer:

VikaD [51]4 years ago

7 0

Answer:

The correct answer is B. Analysis

Explanation:

Error bars are part of the statistical analysis in the scientific method. Once the scientist have collected the data, he or she proceed to the data analysis. A very common way of comparing the data variability is to use error bars in ghaphical representations. From these bars, it can be estimated the error of a determination and experimental groups are compared.

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Which of the following types of electromagnetic waves generally has the least energy

sertanlavr [38]

Radio waves have the lowest energy, but the longest wavelength. So radio waves is the answer. Good luck on the rest and message me for more help.

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3 years ago

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For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)

dangina [55]

Answer:

14.4 g of $CO_{2}$ can be produced.

Explanation:

Balanced equation: $2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O$

Molar mass (g/mol)

$C_{4}H_{10}$ 58.12

$O_{2}$ 32

$CO_{2}$ 44.01

So, 9.60 g of $C_{4}H_{10}$ = $\frac{9.60}{58.12}mol=0.165mol$

17.0 g of $O_{2}$ = $\frac{17.0}{32}mol=0.531mol$

According to balanced equation-

2 moles of $C_{4}H_{10}$ produce 8 moles of $CO_{2}$

So, 0.165 moles of $C_{4}H_{10}$ produce $(\frac{8}{2}\times 0.165)mol$ of $CO_{2}$ or 0.660 moles of $CO_{2}$

13 moles of $O_{2}$ produce 8 moles of $CO_{2}$

So, 0.531 moles of $O_{2}$ produce $(\frac{8}{13}\times 0.531)moles$ of $CO_{2}$ or 0.327 moles of $CO_{2}$

As least number of moles of $CO_{2}$ are produced from $O_{2}$ therefore $O_{2}$ is the limiting reagent.

So, maximum amount of $CO_{2}$ that can be formed = 0.327 moles

= $(0.327\times 44.01)g$

= 14.4 g

3 0

3 years ago

3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent wou

juin [17]

2Al + 3Br2 -------------> 2AlBr3

3 g Al = 0.11 mol Al.

6 g Br2 = 0.0375 mole bromine (it is diatomic). 

moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. 

Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
moles of Al left = 0.11 - 0.025 = 0.086

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4 years ago

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr

seropon [69]

Charles’ Law........

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SVETLANKA909090 [29]

I really don't know and it sucks when i need help and people are giving me false answers.

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