All categories

3 years ago

Does the nucleus (protons and neutrons) weigh more than the electrons?

Chemistry

2 answers:

Butoxors [25]3 years ago

3 0

The mass of nucleons (and thus of the nucleus) is roughly 1000 times greater than that of electrons.

vichka [17]3 years ago

3 0

Yes, protons and neutrons weigh much more than electrons. Which is why the mass of an element is the protons and neutrons added together, disregarding the electrons. Their weight is almost negligible.

You might be interested in

Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U

Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

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,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of nitrogen gas = $79mL/s$

$R_2$ = rate of effusion of sulfur dioxide gas = ?

$M_1$ = molar mass of nitrogen gas = 28 g/mole

$M_2$ = molar mass of sulfur dioxide gas = 64 g/mole

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

$(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}$

$R_2=52mL/s$

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0

2 years ago

What is true about energy in an ordinary chemical reaction? Energy is

Lisa [10]

The answer is statement #3.

4 0

2 years ago

Read 2 more answers

How do oceanographers define salinity?

Studentka2010 [4]

The oceanographers define salinity by the number of grams of salt per kilogram of water.

Salinity is illustrated as how much salt is present in the given amount of water. It is dependent upon how much salt is found in the ocean in the specific regions. If there is a sunny environment, it may evaporate an ample amount of water, and leave behind a lot of salt, thus, the water has more salinity.

The salinity of the ocean is usually measured in Practical Salinity Unit (PSU), it is a unit based on the characteristics of seawater conductivity.

3 0

2 years ago

Read 2 more answers

If the initial concentration of a is 0.00540 m, what will be the concentration after 795 s?

zavuch27 [327]

<span>Feb 19, 2014 - The units of k tell you that this is a second order reaction. So, to solve this, you need to use the integrated rate law for a 2nd order reaction: 1/[A] = kt + 1/[A]o 1/[A] = 0.540/Ms (835 s) + 1/0.00640 1/[A] = 607 [A] = 1.65X10^-3 M.</span><span>
</span>

7 0

2 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

2 years ago