What kind of chemical reaction is this?

What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?

Cerrena [4.2K]

Answer: The de broglie wavelength is $2.037 \times 10^{-34} m$ .

Explanation:

Calculate $\lambda = \frac{h}{p}$ as follows.

$\lambda = \frac{h}{p}$

where,

h = plank's constant = $6.6 \times 10^{-34} m^{2} kg/s$

p = momentum = $mass \times velocity$

Putting the values in the formula as follows.

$\lambda = \frac{h}{mass \times velocity}$

= $\frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}$

= $2.037 \times 10^{-34} m$

Thus, the de broglie wavelength is $2.037 \times 10^{-34} m$ .

6 0

3 years ago

Read 2 more answers

How many electrons will a neon (Ne) atom have when it has no charge?

Darya [45]

Answer:

hope it helps you

Explanation:

Once one shell is full, the next electron that is added has to move to the next shell. So... for the element of NEON, you already know that the atomic number tells you the number of electrons. That means there are 10 electrons in a neon atom.

6 0

3 years ago

Night visin cameras are sensitive to energies around 2.21 x 10-19 J. What wavelength of electromagnetic radiation do they use?

defon

Answer:

8.99×10^-7m

Explanation:

The wavelength can be calculated using the expression below

E=hcλ

Where E= energy= 2.21 x 10^-19 J.

C= speed of light= 3x10^8 m/s

h= planks constant= 6.626 × 10^-34 m2 kg / s

E=hcλ

λ= E/(hc)

Substitute for the values

λ=( 2.21 x 10^-19 )/(6.626 × 10^-34 × 3x10^8 )

= 8.99×10^-7m

6 0

2 years ago

The law of constant composition states: All atoms of a given element have a constant composition and are different than atoms of

Airida [17]

Answer:

C

Explanation:

The law proves C. For examples no matter how water you have it will always have a 1:2 ratio of oxygen to hydrogen. :)

7 0

2 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago