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4 years ago

Which models of the atom include a structure that is mostly made of empty space?

1 answer:

8 0

Thomson proposed this idea. Rutherford disproved it when he shot alpha particles at a gold sheet and they bounced back at him, proving that the alpha particles bounced off something aka the nucleus.

D.

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What is the molality of a solution containing 5.0 moles of solute in 3.2 kg of solvent? Round to the nearest tenth.

Setler [38]

N=5.0 mol
m=3.2 kg

C=n/m

C=5.0/3.2≈1.6 mol/kg

5 0

3 years ago

A sample of solid NH4NO3 was placed in an evacuated container and then heated so that it decomposed explosively according to the

dybincka [34]

Answer:

Kp is 2.98 (option c.)

Explanation:

The decomposition is:

NH₄NO₃(s) → N₂O(g) + 2H₂O(g)

Total pressure at equilibrium is 2.72 bar so, in order to determine Kp we need the partial pressure and we only have total pressure.

According to stoichiometry, nitrogen oxide increase by 1, the partial pressure and water vapor, by 2.

Total pressure is: Partial pressure N₂O + Partial pressure H₂O

2.72 bar = X + 2X → X = 2.72 bar / 3 = 0.91 bar

Partial pressure N₂O = 0.91 bar

Partial pressure H₂O = 1.81 bar

We make the expression for Kp = Partial pressure N₂O . (P. pressure H₂O)²

Kp = 0.91 . 1.81² = 2.98

We do not consider the ammonium nitrate, because it is solid

3 0

3 years ago

Read 2 more answers

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

How to write the formulas

DanielleElmas [232]

If you have to write the chemical formula of a simple, binary ionic compound given the name of the compound, you follow a set of three steps. Let's go through them using magnesium chloride as an example. Write the symbols for the cation and the anion: Mg and Cl. Determine the charge on the cation and anion.

6 0

3 years ago

what is the percent by mass concentration of a solution that contains 5.30 grams of salt dissolved in 19.7 g of water

neonofarm [45]

Answer:

0.212

Explanation:

(5.30g) / (5.30g + 19.7g)

4 0

3 years ago