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2 years ago

How did Ernest Rutherford’s experiment relate to J.J. Thomson’s work?

1 answer:

8 0

Thomson realized that the accepted model of an atom did not account for negatively or positively charged particles. Therefore, he proposed a model of the atom which he likened to plum pudding. ... Rutherford with the assistance of Ernest Marsden and Hans Geiger performed a series of experiments using alpha particles.

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A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for

iren [92.7K]

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

Divide the three by the lowest ratio which is 3.22:

C = 3.22 / 3.22 = 1

H = 16.25 / 3.22 = 5

N = 3.22 / 3.22 = 1

So the empirical formula is:

CHN

5 0

2 years ago

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

2 years ago

Which of the following represents the greater concentration?<br> 1. 5 ppm<br> 2. 5%

Elanso [62]

Answer:

Explanation:

3 0

2 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

3 years ago

Read 2 more answers

2. The pressure of the oxygen gas inside a

Flauer [41]

Answer: 4.41 atm

Explanation:

Given that,

Original pressure of oxygen gas (P1) = 5.00 atm

Original temperature of oxygen gas (T1) = 25°C

[Convert 25°C to Kelvin by adding 273

25°C + 273 = 298K

New pressure of oxygen gas (P2) = ?

New temperature of oxygen gas (T2) = -10°C

[Convert -10°C to Kelvin by adding 273

-10°C + 273 = 263K

Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law

P1/T1 = P2/T2

5.00 atm /298K = P2/263K

To get the value of P2, cross multiply

5.00 atm x 263K = 298K x V2

1315 atm•K = 298K•V2

V2 = 1315 atm•K / 298K

V2 = 4.41 atm

Thus, the new pressure inside the canister is 4.41 atmosphere

4 0

2 years ago