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3 years ago

How did Ernest Rutherford change the atomic model?

Chemistry

1 answer:

nadya68 [22]3 years ago

6 0

Answer:

D. He showed that most of an atom's mass was located in the atom's

nucleus.

Explanation:

Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.

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How does sodium become a negative ion of neon

Svetlanka [38]

I don’t know if this works

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3 years ago

For each of the acid–base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutraliz

LiRa [457]

Acid-base neutralization reaction is defined as reaction of acid with base to form corresponding salt and water. Strong acid and base completely neutralize each other.

(a) The acid base neutralization reaction is as follows:

$HCl(aq)+NaOH(aq)\rightarrow H_{2}O(l)+NaCl(g)$

From the above balanced chemical reaction, 1 mol of NaOH completely reacts with 1 mol of HCl. The mass of NaOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of NaOH is 39.997 g/mol, thus,

$n=\frac{4.85 g}{39.997 g/mol}=0.121 mol$

Thus, 0.121 mol of NaOH reacts with same amount of HCl and number of moles of HCl will be 0.121 mol.

Since. molar mass of HCl is 36.46 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.121 mol\times 36.46 g/mol=4.421 g$

Therefore, 4.421 g of HCl completely reacts with 4.85 g of NaCl.

(b) The acid base neutralization reaction is as follows:

$2HNO_{3}(aq)+Ca(OH)_{2}(aq)\rightarrow 2H_{2}O(l)+Ca(NO_{3})_{2}(aq)$

From the above balanced chemical reaction, 1 mol of $Ca(OH)_{2}$ completely reacts with 2 mol of $HNO_{3}$ . The mass of $Ca(OH)_{2}$ is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of $Ca(OH)_{2}$ is 74.093 g/mol, thus,

$n=\frac{4.85 g}{74.093 g/mol}=0.06545 mol$

Thus, 0.06545 mol of $Ca(OH)_{2}$ reacts with $2\times 0.06545 mol=0.13091 mol$ of $HNO_{3}$

Since. molar mass of $HNO_{3}$ is 63.01 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.13091 mol\times 63.01 g/mol=8.25 g$

Therefore, 8.25 g of $HNO_{3}$ completely reacts with 4.85 g of $Ca(OH)_{2}$ .

$H_{2}SO_{4}(aq)+2 KOH (aq)\rightarrow 2H_{2}O(l)+K_{2}SO_{4}(aq)$

From the above balanced chemical reaction, 2 mol of KOH completely reacts with 1 mol of $H_{2}SO_{4}$ . The mass of KOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of KOH is 56.1056 g/mol, thus,

$n=\frac{4.85 g}{56.1056 g/mol}=0.0864 mol$

Thus, 0.0864 mol of KOH reacts with $\frac{0.0864 mol}{2}=0.0432 mol$ of $H_{2}SO_{4}$

Since. molar mass of $H_{2}SO_{4}$ is 98.079 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.0432 mol\times 98.079 g/mol=4.24 g$

Therefore, 4.24 g of KOH completely reacts with 4.85 g of $H_{2}SO_{4}$ .

8 0

3 years ago

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% soluti

Akimi4 [234]

The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution.
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
.8x = 14
x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

6 0

3 years ago

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A container holds 15.0 g of phosphorous gas at a pressure of 2.0 atm and a temperature of 20.0 Celsius. What is the density of t

erik [133]

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We calculate as follows:

PV = nRT
PV = mRT/ Molar mass
m/V = P(molar mass)/RT
Density = P(molar mass)/RT
Density = 2.0 ( 30.97 ) / 0.08206 ( 20 + 273.15) = 2.57 g/L <----First option

5 0

3 years ago

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Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

$Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)$

Moles of calcium nitrate = $\frac{31.3 g}{164 g/mol}=0.1908 mol$

Moles of ammonium fluoride = $\frac{38.7 g}{37 g/mol}=1.046 mol$

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

$\frac{1}{2}\times 1.046 mol=0.523 mol$ calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0

3 years ago