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Brilliant_brown [7]
3 years ago
11

How many grams are in 2 moles of FeH3?

Chemistry
2 answers:
pshichka [43]3 years ago
3 0

Answer:335.07

Explanation:

LenaWriter [7]3 years ago
3 0

Answer:

Pretty sure it's <u>175.82 grams</u>, sorry if it's wrong!

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In redox reactions, _____________ occurs when electrons are lost by a molecule. ______________ occurs when electrons are gained
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Answer:

Oxidation, Reduction

Explanation:

A redox reaction is a short form for reduction-oxidation.

Reduction is a term which means that electron is gained while oxidation is a term which means that electron Is lost.

The species that gain electron is known as the oxidizing agent while the species losing electrons is known as the reducing agent

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kozerog [31]

Answer: A

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Is a mushroom a heterotroph or autotroph
nadezda [96]

Answer:

A mushroom is a heterotroph.

Explanation:

Mushrooms are fungi, which are heterotrophs because they depend on other organisms for their food.

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Which base would not effectively deprotonate acetylene? (ch3)2nli ch3och2mgbr lioch3 ch3li kh?
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Acetylene has a chemical formula which can be written as:

C2H2

 

We can see that there are two positive ions, H+. Now what deprotonation means is that the H+ is removed from acetylene to form acetylene ion and water. In this case, I believe that the answer would be:

<span>LiOCH3</span>

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Vitamin C has the formula CxHyOz. You burn 0.486 g of the compound in a combustion analysis chamber and isolate 0.728 g of CO2 a
-Dominant- [34]

Answer:

C_3H_4O_3

Explanation:

Hello there!

In this case, according to the given information derived from the combustion analysis, it turns out possible for us to realize that all the carbon comes from the CO2 and all the hydrogen from the H2O, it means we can calculate their moles in the vitamin C as shown below:

n_C=0.728gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.0165molC\\\\n_H=  0.198gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.0220molH

Next, we calculate the grams and moles of oxygen from the grams of C and H in the sample:

m_O=0.486g-0.0165molC*\frac{12.01gC}{1molC} -0.0220molH*\frac{1.01gH}{1molH}\\\\m_O=0.266gO\\\\n_O=0.266gO*\frac{1molO}{16.00gO} =0.0166molO

Then, we divide the moles of C, H, O by 0.0165 as the fewest moles in order to calculate the correct mole ratios:

C=\frac{0.0165}{0.0165}= 1\\\\H=\frac{0.0220}{0.0165}= 1.33\\\\O=\frac{0.0166}{0.0165}= 1

Finally, we turn them into whole number by multiplying by 3 so that the empirical formula is:

C_3H_4O_3

Regards!

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3 years ago
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