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lakkis [162]
3 years ago
12

What was concluded about the structure of the atom as the result of the gold foil experiment?

Chemistry
1 answer:
disa [49]3 years ago
8 0
It's 2,, As a result of the gold foil experiment, A <span>positively charged nucleus is surrounded by mostly empty space.</span>
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How would you prepare a 0.1M solution of HCl starting with a 1.0M solution? Assume you want to prepare 100 ml of solution.
omeli [17]
This question is a dilution problem. In order to prepare a certain concentration and volume from a stock solution, you have to use the equation,

M1V1 = M2V2

From the equation, you need to calculate the volume you need from the stock solution to be able to have the diluted concentration then dilute it to a certain volume.

(1.0 M) (V1) = (0.1 M) (100 mL)
V1 = 0.1 mL is needed from the 1.0 M HCl solution

4 0
3 years ago
Why is carbone reduction process not applied for the extraction of alkali
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Carbon cannot reduce sodium because sodium is stronger than carbon. reactive metals like sodium and potassium and electrolysis has to be used.
7 0
3 years ago
Help anyone can help me do this question,I will mark brainlest.​
Anettt [7]

Answer:

12 grams

Explanation:

if  you used the transformation equation  you would end up with 12 grams

4 0
3 years ago
WILL MARK BRAINLIEST
tiny-mole [99]

Answer:

b) 590

Explanation:

5 0
3 years ago
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so
zmey [24]

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 50ml+50ml=100ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 100 g

T_{final} = final temperature of water = 27.5^0C

T_{initial} = initial temperature of metal = 21.0^0C

Now put all the given values in the above formula, we get:

q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C

q=2719.6J=2.72kJ

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of HCl:

0.05 moles of HCl releases heat = 2.72 KJ

1 mole of HCl releases heat =\frac{2.72}{0.05}\times 1=54.4KJ

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

6 0
3 years ago
Read 2 more answers
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