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3 years ago

Which model of the atom was used as a result of JJ Thomson's cathode ray

Chemistry

1 answer:

mariarad [96]3 years ago

6 0

Answer:

a. Plum pudding model

Explanation:

The plum pudding model of the atom was proposed by J.J. Thomson. It was the model he derived from his experiment on the gas discharge tube.

J.J Thomson was the first person to discover electrons which he called cathode rays because in the discharge tube, they emanate from the cathode.

This led him to suggest the plum pudding model of the atom.
The model reflects electrons being surrounded by a volume of negative charges.

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How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has

Papessa [141]

The balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas= $\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.

number of moles of sodium azide = $16 moles$

Mass of sodium azide in g = $number of moles\times molar mass of sodium azide$ .

= $16 moles\times 65.00 g/mol$

= $1040 g$

Thus, mass of sodium azide which is required to produce $19.0 ft^{3}$ of nitrogen gas = $1040 g$

3 0

3 years ago

Which of the following is a scientific question you could ask about the game

aniked [119]

Answer:

Answer is option C

Explanation:

If you like my answer than please mark me brainliest thanks

3 0

3 years ago

How many grams of oxygen can be obtained from 250. grams of KCIO3?

guapka [62]

Answer:

Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.

Explanation:

3 0

3 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

6 0

4 years ago

For the following reaction, identify whether the compound in bold is behaving as an acid or a base.

harina [27]

Answer: acid

Explanation: it dissolves in water to produce hydronium ion as the only positive ion

3 0

4 years ago