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Morgarella [4.7K]

3 years ago

11

After James Chadwick discovered the neutron, atomic theory expanded to include isotopes. Isotopes have many practical uses. For

example,
scientists use the relative amount of carbon-12 to carbon-14 in ancient bones to determine how old they are.

How many neutrons are in an atom of Carbon-147
A 2
B 6
C 8
D 14

1 answer:

andrew-mc [135]3 years ago

8 0

Answer:

The answer is C which is 8 neutrons

Explanation:

This is because all carbon isotopes have 6 protons in their nucleus. Carbon 14 have 2 extra neutron aside from only having 6. So 6 protons + 6 neutrons + 2 more extra makes 14 which makes it Carbon - 14. I hope this makes sense.

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Tcecarenko [31]

Ratio of neutrons to protons and <span>number of electrons in the outer shell</span>

4 0

3 years ago

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Which two practices are examples of how people use science?

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B. Developing theories using many lines of evidence

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The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

7 0

3 years ago

Given the reaction agi(s) ↔ ag+(aq) + i-(aq) solution equilibrium is reached in the system when

Julli [10]

Missing question:
1) the rate of dissolving reaches zero
<span>2) the rate of crystallization reaches zero </span>
3) the rate of dissolving is zero and the rate of crystallization is greater than zero.
<span>4) both the rate of dissolving and the rate of crystallization are equal and greater than zero.
</span>
Answer is: 4) both the rate of dissolving and the rate of crystallization are equal and greater than zero.
Silver chloride (AgCl) dissolves and form silver and chlorine ions, in the same time silver and chlorine ions crystallizate and form solid salt silver chloride.
In equilibrium rates of dissolvinf and crysallization and concentration of ions do not change.

7 0

3 years ago

The fire department needs information on friction losses occurring between a water main and an open fire hydrant. At maximum mai

lisov135 [29]

Explanation:

The given data is as follows.

$P_{1}$ = 85 psig, $P_{2}$ = $P_{atm}$ = 15 psia

Q = 1620 gpm, d = 2.5 inch, l = 8 ft = 2.4384 m

According to Darey-Weisbach equation,

$h_{l} = \frac{4fl \nu^{2}}{2gD}$ ......... (1)

Value of 'f' will be decided on the basis of Reynold number.

As, it is known that $R_{l} = \frac{\rho \nu d}{\mu}$

where, $\mu_{water}$ = $10^{-3}$ kg/ms

As it is known that 1 gpm = $\frac{1}{3.67} m^{3}/hr$

So, $1 m^{3}/hr$ = 3.67 gpm

Therefore, Q = $1620 \times \frac{1}{3.67}$

= $441.4168 m^{3}/hr$

= 0.1226 $m^{3}/s$

In, 1 inch = 2.54 cm = 0.0254 m

Therefore, d = 2.5 \times 0.0254 = 0.0635 m

V = $\frac{Q}{\frac{\pi}{4}d^{2}}$

= $\frac{0.1226}{0.785 \times (0.0635)^{2}}$

= 38.73 m/s

Hence, we will calculate Reynold number as follows.

$R_{l}$ = $\frac{1000 \times 38.73 \times 0.0635}{10^{-3}}$

= 2459355

As $R_{l}$ > 2000 then, it means that flow is turbulent.

As, f = 0.079 $R^{-0.25}_{l}$

= 0.001994

Putting all the values into equation (1) formula as follows.

$h_{l} = \frac{4fl \nu^{2}}{2gD}$

= $\frac{4 \times 0.001994 \times 2.4384 \times (38.73)^{2}}{2 \times 9.81 \times 0.0635}$

= $1.04069 \times 10^{5} m$

Thus, we can conclude that friction loss from the main to the discharge point is $1.04069 \times 10^{5} m$ .

4 0

3 years ago