How many atoms are present in 179.0 g of iridium?

Fter a radioactive atom decays, it is the same element that it was before with no measurable change in mass. Which kind of decay

Alborosie

After a radioactive atom decays, it is the same element that it was before with no measurable change in mass. the decay that is present is gamma decay because gamma decay has photons which has no mass unlike alpha and beta decay.

6 0

2 years ago

Read 2 more answers

A newspaper story describing the local celebration of Mole Day on October 23 (selected for Avogadro's number, 6.02 X 10^ 23 ) at

stepladder [879]

Answer:

The answer is "670.176 km".

Explanation:

Volume of the occupies of one M & M= 0.5 \ cm^3\\\\

M&M 1 mole $= 6.02 \times 10^{23}\\\\$

Calculating volume of M& M mole $= 0.5 \ cm^3 \times 6.02 \times 10^{23}=3.01 \times 10^{23}\ cm^3\\\\$

Calculating the cube mole $L^3= 3.01 \times 10^{23}\ cm^3\\\\$

$L=6.70176 \times 10^{7}\ cm\\\\$

$=6.70176 \times 10^{7}\ cm \times \frac{1\ m}{100\ cm}\\\\=670176\ m\\\\=670.176 \ km$

Therefore 18 tractor trailers wouldn't be sufficient.

8 0

3 years ago

CAN YOU PLEASE HELP ASAP

Helga [31]

Your answer would be D you welcome !

7 0

2 years ago

Read 2 more answers

As you get closer to the boundary between two of the Earth's Plates what is most likely to happen to the intensity of the Earthq

Zigmanuir [339]

The Ring of fire most seismically and volcanically active zone in the world.

4 0

2 years ago

Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0.2

Allushta [10]

This is an incomplete question, here is a complete question.

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse.

The chemical reaction is:

$XY(g)\rightleftharpoons X(g)+Y(g)$

Concentration(M) [XY] [X] [Y]

(M)initial: 0.200 0.300 0.300

change: +x -x -x

equilibrium: 0.200+x 0.300-x 0.300-x

The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a Kc value of 0.140 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answer : The Equilibrium concentrations of XY, X, and Y is, 0.104 M, 0.204 M and 0.204 M respectively.

Explanation :

The chemical reaction is:

$XY(g)\rightleftharpoons X(g)+Y(g)$

initial: 0.200 0.300 0.300

change: +x -x -x

equilibrium: (0.200+x) (0.300-x) (0.300-x)

The equilibrium constant expression will be:

$K_c=\farc{[X][Y]}{[XY]}$

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

$0.140=\frac{(0.300-x)\times (0.300-x)}{(0.200+x)}$

By solving the term 'x', we get:

x = 0.0963 and x = 0.644

We are neglecting the value of x = 0.644 because the equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.0963

Equilibrium concentrations of XY = 0.200+x = 0.200+0.0963 = 0.104 M

Equilibrium concentrations of X = 0.300-x = 0.300-0.0963 = 0.204 M

6 0

3 years ago