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N76 [4]
2 years ago
12

Give the main conclusion from the 1911

Chemistry
1 answer:
Monica [59]2 years ago
3 0

Answer:

Here's what I get  

Explanation:

From the scattering experiments, Rutherford concluded that the positive charge was concentrated in the nucleus of the atom.

This contradicted the plum pudding model, which assumed that the positive charge was uniformly distributed throughout the atom.

You might be interested in
What is SO2 shape name?
muminat

Answer:Molecular Formula SO2

Hybridization Type sp2

Bond Angle 119o

Geometry V-Shaped or Bent

Explanation:

hope this helped <3

5 0
3 years ago
An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the am
gogolik [260]

Answer:

Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

Explanation:

4 0
2 years ago
Metric conversions.<br> Please help ASAP.
lesya692 [45]

Answer:

14. 13.2cg = 1.32dg

15. 3.8m = 0.0038km

16. 24.8L = 24800mL

17. 0.87kL = 870L

18. 26.01cm = 0.0002601km

19. 0.001hm = 10cm

Explanation:

14. 13.2/10 = 1.32

15. 38/1000 = 0.0038

16. 24.8(1000) = 24,800

17. 0.87(1000) = 870

18. 26.01/100000 = 0.0002601

19. 0.001hm(10000) = 10

An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.

7 0
3 years ago
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
Svet_ta [14]

Answer:

V_1=23.3~mL

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

C_1*V_1=C_2*V_2

Now we can identify the variables:

C_1=~1.475_M

V_1=~?

C_2=~0.1374_M

V_2=~250.0~mL

If we plug all the values into the equation:

1.475_M*V_1=0.1374_M*250.0~mL

And we solve for V_1:

V_1=\frac{0.1374_M*250.0~mL}{1.475_M}

V_1=23.3~mL

I hope it helps!

8 0
2 years ago
A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

3 0
3 years ago
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