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xeze [42]
3 years ago
12

A student made an initial 1:5 dilution of protein lysate. Then 2mL of that was added to 8mL of water. Lastly, the student made a

1:20 dilution of the second tube. What is the final dilution of protein lysate.
Chemistry
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

The final dilution is 1:400

Explanation:

Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.

So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400

This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.

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What is the electron configuration of the element in period 2 that has 5 valence electrons (valence electrons are the electrons
balu736 [363]

Answer:

Nitrogen

Explanation:

N= is the symbol

electronic configuration

1s2 2s2 2p3

8 0
3 years ago
Mn2+(aq) + nabio3(s) → bi3+(aq) + mno4−(aq) + na+(aq) how many water molecules are there in the balanced equation (for the react
stiks02 [169]
<span>
Mn²⁺ + 4H2O -----> MnO4⁻ + 8H⁺ +5e⁻                  /*2
<span>NaBiO3 +6H⁺ +2e⁻ -----> Bi³⁺ + Na⁺ + 3H2O        /*5
</span>2Mn²⁺ + 5 NaBiO3+8H2O+30H⁺ --->  2MnO4⁻ +5Bi³⁺ + 5Na⁺ +16H⁺ +15H2O

</span>2Mn²⁺ + 5 NaBiO3+14H⁺ --->  2MnO4⁻ +5Bi³⁺ + 5Na⁺  +7H2O

There are 7 water molecules in this reaction.
4 0
3 years ago
Read 2 more answers
Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt% methanol and the second contain
RoseWind [281]

Answer:

M_{per}= 52.86

W_{per}=47.14

Explanation:

<u>First mixture</u>:

40 wt% methanol - 60 wt% water 200 kg

m_{met1}=200 kg * 0.4= 80 kg

m_{wat1}=200 kg * 0.6= 120 kg

<u>Second mixture</u>:

70 wt% methanol - 30 wt% water 150 kg

m_{met2}=150 kg * 0.7= 105 kg

m_{wat2}=150 kg * 0.3= 45 kg

Final mixture:

m_{metF=80 kg + 105 kg= 185 kg

m_{watF}=120 kg + 45 = 165 kg

M_{per}=\frac{185 kg}{185 kg + 165 kg}*100= 52.86

W_{per}=\frac{165 kg}{185 kg + 165 kg}*100=47.14

If, the compositions are constant, the only variables are the mass of each mixture used in the final one, so there can be only one independent balance.

8 0
3 years ago
the hemiacetal below is treated with 18o-labeled methanol (ch3o*h) and acid. where will the label appear in the products?
ioda

The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

<h3>Position of 18o-labeled methanol in the products</h3>

The 18O label will appear at position b in the product as indicated in the image.

This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).

While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.

Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

Learn more about methanol here: brainly.com/question/17048792
#SPJ11

8 0
2 years ago
Cu is in FCC structure. We know that even at close to melting temperature, vacancies are still sparsely distributed, i.e., it is
sveticcg [70]

Answer:

6

Explanation:

FCC is face centered cubic lattice. In FCC structure, there are eight atoms at the eight corner of the cubic unit cell and one atom centered in each of the faces. FCC unit cells consist of four atoms, (8/8) at the corners and (6/2) in the faces.

Given that, Cu has FCC structure and it contains a vacancy at origin (0, 0, 0). And there is no other vacancy directly adjacent to the vacancy at the origin. So, all the adjacent positions contain Cu atoms. Hence, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy.

the above FCC unit cell clearly indicates that there are six adjacent atoms adjacent to the vacancy at origin

So, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy is 6.

5 0
3 years ago
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