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lana [24]
3 years ago
14

Please following me in the branily app ~♥~​

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

Answer:

\huge\colorbox{pink}{✏﹏ \: BlackPink\: }

Omg!!!!! BlackPink!!!!!

# Blinkeu~~ ♡

Nonamiya [84]3 years ago
3 0
If their was a follow button on a profile why not I would but their isn’t sooo I mean I’ll follow these convos so what’s up hope everyone having a good day :)
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What kind of radioactive element is useful for dating an object?
OlgaM077 [116]

The kind of radioactive element is useful for dating an object is one with a half-life close to the age of the object.

<h3>What are radioactive elements?</h3>

Radioactive elements are elements that involved in radioactivity and this comprises of atoms or particles in their nuclei whose atomic nuclei are not stable and they emit radiations to maintain stability.

Therefore, The kind of radioactive element is useful for dating an object is one with a half-life close to the age of the object.

Learn more about radioactive elements below.

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4 0
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Chlorine released from chlorofluorocarbons (cfcs) reacts with ozone in the atmosphere to form oxygen. the proposed reactions are
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The catalyst is what appears exactly the same at the end and appears early in the equation set. In this case Cl(g).The intermediate appears "intermediately" not at the beginning or at the end, but is made and consumed in the middle. Like ClO(g). A substance that is regenerated in the next is a catalyst and is consumed in the first step. In contrast, when a substance is formed in the first step and is consumed in the next step, then it is known as an intermediate.
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The Great Burdock plant’s seeds have spines on them that attach to the fur of animals that brush against it. The seed then trave
Sphinxa [80]
The answer is B. Commensalism.

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Butane reacts with oxygen to produce carbon dioxide and water
Wittaler [7]
What exactly are you looking for?
This is the balanced equation.
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4 0
4 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
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