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2 years ago

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If the experiment were repeated using a cuvettes with a path length of 2.0 cm, how would that affect the absorbance values for t

he Co2 solutions (if at all), assuming all of the concentrations remained unchanged?

1 answer:

inessss [21]2 years ago

5 0

Answer:

..............................

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As a human being, we are considered to be a ..... because we rely on other organism for food

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Heterotrophs i think :)

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Convert 16.8 L of nitrogen monoxide gas at STP to grams .

german

Answer:

See explanation

Explanation:

1 mole of a gas occupies 22.4 L

x moles occupies 16.8 L

x = 1 mole * 16.8 L/22.4 L

x = 0.75 moles

number of moles = mass/molar mass

mass = number of moles * molar mass

mass = 0.75 moles * 30.01 g/mol = 22.5075 g = 2.25 * 10^1 g

the coefficient of the scientific notation answer = 2.25

the exponent of the scientific notation answer = 1

significant figures are there in the answer = 6

the right most significant figure in the answer = 3

2.

number of moles = 12.5g/38g/mol = 0.3289 moles

1 mole occupies 22.4 L

0.3289 moles occupies 0.3289 moles * 22.4 L/1 mole

= 7.36736 L = 7.36736 * 10^0 L= 7.37 * 10^0 L

the coefficient of the scientific notation answer =7.37

the exponent of the scientific notation answer = 0

significant figures are there in the answer = 6

the right most significant figure in the answer= 3

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2 years ago

Water covers over 70% of planet Earth and most of the water is salt water. Which

Serga [27]

Answer:

D. 4

Explanation:

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Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo

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Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

$n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4$

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

$left\ over=0g$

Regards.

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2 years ago

Which of these elements is the most ductile?

Inessa05 [86]

The answer is going to be mercury

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