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Arte-miy333 [17]
3 years ago
13

What is the m∠L? 45° 54° 40° 52°

Chemistry
1 answer:
Nataly [62]3 years ago
4 0

Answer:

m<L = 45

Explanation:

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Identify which statement describes the following process. 2Ce + 3Cu2+ → 3Cu + 2Ce3+
Georgia [21]
Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.

</span>There are two reactions in the equation, oxidation and reduction. A molecule that oxidized will lose electrons while the molecule that reduced will gain electrons. In this case, Cu2+ changed into Cu which means its oxidation number reduced from +2 into 0. Ce oxidation number increased from 0 into +3 
4 0
3 years ago
A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What i
spayn [35]

Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm

8 0
3 years ago
For the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to
stira [4]
This may help you
<span>You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass. Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced. To go back to grams, multiply the moles of each product that you get by their molar masses!</span>
7 0
4 years ago
HOLMIUM
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Answer:

1. The graph where x axis and y axis are present is called coordinate.

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8 0
3 years ago
How many grams of carbon are present in 45.0 g of CCl4?
lisabon 2012 [21]
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

Hope this answers the question. Have a nice day.
5 0
3 years ago
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