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densk [106]
3 years ago
13

In the electrolysis of molten coi3, which product forms at the cathode? 1. i2(g) 2. h2(g) 3. o2(g) 4. co(l)

Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0
THE ANSWER TO THIS QUESTION WILL BE THE FOURTH OPTION Co(l)
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What happens to the population size of each of the pond organisms when the perch population size is decreased? How many of the s
lana [24]
I found a presentation of Food web of a pond that will greatly connect with the above problem.  http://www.eduweb.com/portfolio/earthsystems/food/foodweb4.html

Normal setting:
Species        Pop. size
Blue heron   Medium
Perch            Medium
Bass              Medium
Minnows      Medium
Inverts          Medium
Algae           Medium

If PERCH population size decreases, 
Species        Pop. sizeBlue heron   LowPerch            LowBass              MediumMinnows      HighInverts          HighAlgae           Low

As you can see, 4 other species are affected when the Perch population size decreased. Bass is not affected. 
8 0
3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
Questions for buffers lab, please help.
TEA [102]

Answer:

because the acid properties of aspirin may be problematic.

4 0
2 years ago
What is the Definition of "Folded Mountain" in science terms?
just olya [345]
Fold mountains<span> are </span>mountains<span> that form mainly by the effects of </span>folding<span> on layers within the upper part of the Earth's crust. Before either plate tectonic theory developed, or the internal architecture of thrust belts became well understood, the term was used for most</span>mountain<span> belts, such as the Himalayas.</span>
3 0
3 years ago
If 25 mol of C8H18 are available, how many mol of CO2 can be produced
Nady [450]

Answer:

200 mol

Explanation:

2C8H18+25O2=16CO2+18H2O

2 mol → 16 mol

25 mol → X mol

X=25×16÷2=200

7 0
3 years ago
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