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AfilCa [17]
3 years ago
13

How many centimeters are in 5 miles? Hint: 1 miles = 5280 feet

Chemistry
1 answer:
KatRina [158]3 years ago
7 0
804,672 centimeters ❤️‍
You might be interested in
Which statement about chemical reactions is not true?
igomit [66]

Answer:

1: New atoms are formed as products

Explanation:

matter (atoms) cannot be created or destroyed

unless you are God or Cinderella's fairy Godmother or something

7 0
3 years ago
If you obtain 3.0 grams of aspirin from an experiment that could make no more than 3.14 grams, what is the percent yield?
Masteriza [31]

Answer:

96%

Explanation:

To find the percent yield, we can use this equation

\frac{Actual}{Theoretical} *100

The actual yield of aspirin is 3.0 and the theoretical is 3.14 in this case, so just plug the numbers in.

\frac{3.0}{3.14} *100\\\\ =96

Thus the percent yield is 96%

;)

3 0
3 years ago
A student assisting with the experiment would observe all of the following about the electron transport chain EXCEPT:A. Electron
zheka24 [161]

Answer:

C. All electron carriers are mobile and hydrophobic

Explanation:

Hello,

In this case, it is widely known that the electron carriers move inside the inner mitochondrial membrane and consequently move electrons from one to another. In such a way, they are mobile, therefore they are largely hydrophobic as long as they are inside the membrane.

For instance, the cytochrome c is a water-soluble protein in a large range, therefore, the answer is: C. All electron carriers are mobile and hydrophobic.

Best regards.

5 0
3 years ago
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

3 0
3 years ago
The new boiling point is 101.02°C. Suppose that you add another 2.0 mol of sucrose to this solution. What do you predict the new
solniwko [45]

Mad Lad, thanks for the answer :)

7 0
3 years ago
Read 2 more answers
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