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Norma-Jean [14]

3 years ago

8

If the common element of Fluorine-9 and one of the isotopes is Fluorine-10, what does the -10 tell us about the element and how

many protons and neutrons does the element have?

1 answer:

guajiro [1.7K]3 years ago

8 0

flourine have 9 proton

19 neutrons and 9 electron

mark be a brain list

You might be interested in

Rutherford’s gold foil experiment provided evidence for which of the following statements?

nignag [31]

The answer is (c.) There is a dense, positively charged mass in the center of an atom
In 1899, Ernest Rutherford, a British physicist conducted a study of the absorption of radioactivity by thin sheets of metal foil. On this experiment, he concluded that all the positive charge and the mass of the atom is concentrated in a small fraction of the total volume of the atom which is called the Nucleus.

7 0

4 years ago

Determine the number of protons and neutrons in plutonium-239 and enter its symbol in the form azx.

Komok [63]

Answer:

number of protons: 94
number of neutrons: 145

239
       Pu
   94

Explanation:

1) The atomic number, Z, of plutonium is Z = 94, i.e plutonium 94 protons

2) Plutoniun-239 is the isotope with mass number 239.

3) Mass number = number of protons + number of neutrons =>

number of neutrons = mass number - number of protons = 239 - 94 = 145

4) The notation requires that you indicate the symbol of the element with the atomic number (number of protons) and the mass number.

You put the mass number as a superscript at the left side of the symbol and the atomic number as subscript to the left of the symbol.

So, in this case the symbol is Pu, the superscript to the left is 239 and the subscript to the left is 94.

239
       Pu
   94

7 0

3 years ago

In 2.00 min, 29.7 mL of He effuse through a small hole. Under the same conditions of pressure and temperature, 9.28 mL of a mixt

sergiy2304 [10]

Answer : The percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

And the relation between the rate of effusion and volume is :

$R=\frac{V}{t}$

or, from the above we conclude that,

$(\frac{V_1}{V_2})^2=\frac{M_2}{M_1}$ ..........(1)

where,

$V_1$ = volume of helium gas = 29.7 ml

$V_2$ = volume of mixture = 9.28 ml

$M_1$ = molar mass of helium gas = 4 g/mole

$M_2$ = molar mass of mixture = ?

Now put all the given values in the above formula 1, we get the molar mass of mixture.

$(\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}$

$M_2=40.97g/mole$

The average molar mass of mixture = 40.97 g/mole

Now we have to calculate the percent composition by volume of the mixture.

Let the mole fraction of $CO$ be, 'x' and the mole fraction of $CO_2$ will be, (1 - x).

As we know that,

$\text{Average molar mass of mixture}=\text{Mole fraction of }CO$

$\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)$

Now put all the given values in this expression, we get:

$40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)$

$x=0.1894$

The mole fraction of $CO$ = x = 0.1894

The mole fraction of $CO_2$ = 1 - x = 1 - 0.1894 = 0.8106

The percent composition by volume of mixture of $CO$ = $0.1894\times 100=18.94\%$

The percent composition by volume of mixture of $CO_2$ = $0.8106\times 100=81.06\%$

Therefore, the percent composition by volume of mixture of $CO$ and $CO_2$ are, 18.94 % and 81.06 % respectively.

6 0

3 years ago

Need a point ,...pick an answer

sammy [17]

All of them oklol jklllig

4 0

3 years ago

Read 2 more answers

over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th

Aleonysh [2.5K]

Answer : The moles of methane gas could be, $7.90\times 10^{-3}mol$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$

$[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = rate of effusion of fluorine gas

$R_2$ = rate of effusion of methane gas

$n_1$ = moles of fluorine gas = $5.13\times 10^{-3}mol$

$n_2$ = moles of methane gas = ?

$t_1=t_2$ = time = 12.3 min (as per question)

$M_1$ = molar mass of fluorine gas = 38 g/mole

$M_2$ = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

$[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}$

$n_2=7.90\times 10^{-3}mol$

Therefore, the moles of methane gas could be, $7.90\times 10^{-3}mol$

8 0

3 years ago