All categories

3 years ago

How does atomic mass differ from atomic number

Chemistry

2 answers:

valentinak56 [21]3 years ago

5 0

One is weight and one is amount

Lina20 [59]3 years ago

3 0

The atomic mass is the average mass of all the isotopes. The atomic number is the number of protons in the nucleus of an atom. In an uncharged atom the atomic number is also equal to the number of electrons.

Hope this helps!

You might be interested in

Give the quantum number set for one electron in the 3p sublevel of a sulfur (S) atom.

IceJOKER [234]

<span>sulfurs electronic configuration is 1s2 2s2 2p6 3s2 3p4 so for the 3p sublevel principal quantum no - 3 angulum quantum no - 1 magnetic quantum no - (-1) or 0 or (-1) spin quantum no - (+1/2) or (-1/2) so the quantum numbers will be [ 3 , 1, -1 , +1/2 ] or [ 3 , 1, -1 , -1/2 ] [ 3 , 1, 0 , +1/2 ] or [ 3 , 1, 0 , -1/2 ] [ 3 , 1, 1 , +1/2 ] or [ 3 , 1, 1 , -1/2 ]
Hope This Helps</span>

7 0

3 years ago

At a party, you see someone on the floor with grayish skin, having

denis-greek [22]

All of above. Overdoses have side effects like troubles breathing and other

6 0

3 years ago

Read 2 more answers

The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2

notsponge [240]

Answer : The unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law expression for the reaction:

$\text{Rate}=k[N_2]^a[H_2]^b$

where,

a = order with respect to $N_2$

b = order with respect to $H_2$

Expression for rate law for first observation:

$0.0021=k(0.10)^a(0.10)^b$ ....(1)

Expression for rate law for second observation:

$0.0084=k(0.10)^a(0.20)^b$ ....(2)

Expression for rate law for third observation:

$0.0672=k(0.20)^a(0.40)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2$

Dividing 3 by 1 and also put value of b, we get:

$\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[N_2]^a[H_2]^b$

$\text{Rate}=k[N_2]^1[H_2]^2$

Now, calculating the value of 'k' by using any expression.

$0.0021=k(0.10)^1(0.10)^2$

$k=2.1M^{-2}min^{-1}$

The value of the rate constant 'k' for this reaction is $2.1M^{-2}min^{-1}$

That means, the unit for the rate constant in the rate law for the formation of ammonia is, $M^{-2}min^{-1}$

7 0

3 years ago

a given sample of a gas has a volume of 3.0 L at a pressure of 4.0 atm. If the temperature remains constant and the pressure is

TEA [102]

According to boyle p1v1=p2v2
So 3*4=6*x so x=2

6 0

3 years ago

How Do Glow Sticks Glow?

Gekata [30.6K]

Answer:

All liquid glow products depend on a chemical process known as CHEMILUMINESCENCE to produce their light. Chemiluminesence is a chemical reaction that causes a release of energy in the form of light. To produce this light the electrons in the chemicals become excited and rise to a higher energy level.

To utilise this process glowsticks contain two liquids; hydrogen peroxide and tert-butyl alcohol. When mixed together it is these liquids that create the glow. Fluorescent dyes are also used in the alcohol to alter the colour of the light emitted.

Explanation:

8 0

3 years ago