All categories

3 years ago

Which letter indicates the wavelength of the wave?

Chemistry

2 answers:

Anuta_ua [19.1K]3 years ago

7 0

Answer: Letter D represents the wavelength of the wave.

Explanation: We are given a wave in the image. Wave comprises of crests and troughs.

Wavelength of a wave is defined as the distance between the two crests or two troughs or one crest and one trough.

And here letter D is the distance between the two troughs. Hence, this indicates the wavelength of the wave.

Svet_ta [14]3 years ago

3 0

D represents the wavelength in your problem. The wavelength is the measurement between two crests and can be measured on the bottom crest or the top crest.

You might be interested in

How does concentration affect the chemical equilibrium?

nevsk [136]

When the concentration of a reactant is increased, the chemical equilibrium will shift towards the products. More product is formed and the concentration of the reactants decreases as the concentration of the products increases.

7 0

3 years ago

Please help Asap………………

vladimir2022 [97]

Answer:

3rd

Explanation:

6 0

3 years ago

Read 2 more answers

How can one describe the bond between the two atoms of hydrogen to the one atom of oxygen within a molecule of water (h2o)?

Digiron [165]

There are O-H bonds in H2O. They have the intramolecular force of polar covalent bond.

4 0

3 years ago

What is the molar mass of h2

alekssr [168]

Answer:

The answer is one

Explanation:

I will just type any rubbish here bcuz my answer should be more than 20 words..... Just know the answer is 1

3 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$