If the strength of the magnetic field at B is 20 units, the strength of the magnetic field at A is

Why are scientific models important in the study of science? A. They are useful only for teaching students about science. B. The

madreJ [45]

The correct answer should be B.
The correct answer would be B.

4 0

3 years ago

Read 2 more answers

What is a period on the Periodic Table? *

ella [17]

Answer:

Explanation:En la tabla periódica los elementos están ordenados de forma que aquellos con propiedades químicas semejantes, se encuentren situados cerca uno de otro. Los elementos se distribuyen en filas horizontales, llamadas períodos.

8 0

2 years ago

Would it be C..? <br> More NO2 and SO2 would form..help..please and thank you

Scorpion4ik [409]

Answer:

C. More NO2 and SO2 will form

Explanation:

Le Chatelier's Principle : It predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc

It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.

In given Reaction SO3 is introduced(increased) .

So equilibrium will shift in the direction where SO3 should be consumed(decreased)

Hence the equilibrium will go in backward direction , i.e

$NO + SO_{3} \rightarrow NO_{2} + SO_{2}$

So more and more Of NO2 and SO2 will form

7 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 $(\Delta S^o)$ .

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

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

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

How can you distinguish between crystalline allotropic modifications of Sulphur from those of amorphous allotrops?

Burka [1]

The crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

<h3>What is a crystalline substance?</h3>

A crystalline substance is one that has a definite arrangement of the atoms in the substance. An amorphous substance lacks this definite arrangement. We can see this arrangement when we conduct an X-ray crystallography of the sulfur.

Also, the crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

Learn more about sulfur:brainly.com/question/13469437

#SPJ1

4 0

2 years ago