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2 years ago

Which of Hubble's findings supported the big bang theory? Check all that apply.

Chemistry

2 answers:

Stella [2.4K]2 years ago

8 0

Answer:

C,D,E

Explanation:

JUST TOOK THE TEST

grigory [225]2 years ago

3 0

Out of all given options, Hubble's findings supported the big bang theory are as follows: 

The universe formed from a central point.
Most galaxies in the universe are moving away from Earth.
There are billions of galaxies in the universe, not just one.

Option C, D, and E

Explanation:

According to big bang theory, the universe is increasing its region of occupancy in free space. As it is known that the universe is composed of several galaxies. So this spreading of the universe is nothing but spreading of galaxies.

And this is confirmed from Hubble’s findings using Hubble’s telescope. As they have observed that the speed at which the galaxies are travelling in free space has been increased at a faster rate.

So, the galaxies are used to move away from Earth with time at a faster rate which confirms the big bang theory of universe formation. Also, in simple words, universe started from big bang according to cosmology' standard theory, it’s just moves from everything else.

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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

Natasha_Volkova [10]

Answer:

Explanation:

6 0

2 years ago

Read 2 more answers

determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4

Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

HNO₂ ⇄ H⁺ + KNO₂⁻

C(i) 0.55M 0M 0.75M

ΔC -x +x +x

C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be

dropped giving ...

≅0.55M x ≅0.75M

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0

2 years ago

Experiments were carried out in which a beam of cathode rays was first bent by a magnetic field and then bent back by an electro

Sergio039 [100]

a. the ratio of mass to charge of an electron

Explanation:

The experiment permitted the direct measurement of the ratio of mass to charge of an electron.

The charge to mass ratio of an electron was determine by accelerating a beam of cathode rays in magnetic and electric fields.
No matter the gas used in the tube or the nature of the material of the electrodes, the rays were found to have constant charge to mass ratio of 1.76 x 10¹¹coulombkg⁻¹.

learn more:

Subatomic particles brainly.com/question/2757829

#learnwithBrainly

6 0

3 years ago

Could an organism survive without its mitochondria if all the other organelle were present

lisov135 [29]

No, mitochondria convert the sugar, can give it(the cell) Nutrients (food) to mainly keep it fresh and alive. Without cells mitochondria can't do its job.

8 0

3 years ago

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