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2 years ago

During the Big Bang, what were the two main chemical elements created?

2 answers:

4 0

During the big band only the lightest elements were formed – hydrogen and helium along with trace amounts of lithium and beryllium.

Tasya [4]2 years ago

3 0

this is the answer I got hydrogen and helium

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While soils are an important part of the global carbon cycle, they are not a part of the nitrogen and sulfur cycles.

Vladimir79 [104]

The correct answer is - False.

The soils are part of most of the major cycles that take place on the Earth, mainly because they are in touch with the other spheres. The carbon dioxide, as well as the nitrogen and the sulfur cycles too, end up in the soil in more cases than not during their cycles. While some are formed in it and than released, like the sulfur, the carbon mostly gets in it though the roots of the plants, as well as the decomposing organisms, and the nitrogen ends up in the soil with the water.

The soil is one of the most important pieces in the cycles of most of the gases on Earth, and without it, some will not even be possible.

7 0

2 years ago

Which equation represents a redox reaction?

labwork [276]

Answer:

D. $2NaBr + Cl_2\rightarrow 2NaCl + Br_2$

Explanation:

Hello!

In this case, for the given set of chemical reactions, it is possible to infer that D. is a categorized as redox due to the following:

Since both chlorine and bromine remain as diatomic gases, their oxidation states in such a form is 0, but as anions with lithium cations they have a charge of - according to the following reaction and half-reactions:

$2NaBr + Cl_2\rightarrow 2NaCl + Br_2$

$Cl_2^0+2e^-\rightarrow 2Cl ^-\\\\2Br^- \rightarrow Br_2^0+2e^-$

Unlike the other reactions whereas no change in the oxidation states is evidenced.

6 0

2 years ago

Read 2 more answers

Equivalent massof sodium

andre [41]

Answer:

Sodium

Formula : Na

Equivalent mass:23.0

Mark me as Branliest

8 0

2 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

2 years ago

The electronic configuration of an element is given below.

Shalnov [3]

Answer:

It is reactive because it has to gain an electron to have a full outermost energy level.

Explanation:

The electron configuration of oxygen is 1s2,2s2 2p4.

Oxygen is in group six in the periodic table so it has six electrons in its valence shell. This means that it needs to gain two electrons to obey the octet rule and have a full outer shell of electrons (eight).

3 0

1 year ago