Which one of the following is not a compound?

Why did the big bang not produce a significant proportion of elements heavier than helium?

Travka [436]

The big bang did not produce a significant proportion of elements heavier than helium because the temperatures and densities present in the early universe were not sufficient to support the fusion of heavier elements.

During the first few minute of the big bang, the universe was composed of mostly hydrogen and helium, with very small amounts of lithium and beryllium. As the universe expanded and cooled, the denser regions of the universe collapsed to form the first stars. Inside these stars, the intense pressure and heat generated by nuclear fusion reactions allowed for the production of heavier elements, such as carbon and oxygen. However, elements heavier than helium, such as iron and nickel, require even higher temperatures and densities to be produced, which can only be found in the cores of supernovae. Therefore, the big bang alone did not produce a significant proportion of elements heavier than helium.

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7 0

11 months ago

In a chromatography experiment, chlorophyll pigments are separated using paper. What is the stationary phase in this experiment?

stira [4]

The correct answer is paper

7 0

2 years ago

On a mission to a newly discovered planet, an astronaut finds chlorine abundances of 13.85 % for 35Cl and 86.15 % for 37Cl. What

topjm [15]

Answer : The atomic mass of chlorine is, 36.723 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Mass of 35-Cl isotope = 35 amu

Percentage abundance of 35-Cl isotope = 13.85 %

Fractional abundance of 35-Cl isotope = 0.1385

Mass of 37-Cl isotope = 37 amu

Percentage abundance of 37-Cl isotope = 86.15 %

Fractional abundance of 37-Cl isotope = 0.8615

Now put all the given values in above formula, we get:

$\text{Average atomic mass }=[(35\times 0.1385)+(37\times 0.8615)]$

$\text{Average atomic mass }=36.723amu$

Therefore, the atomic mass of chlorine is, 36.723 amu

6 0

3 years ago

Calculate the Equilibrium constant Kc at 25 C from the free - energy change for the following reaction . Zn(s) + 2Ag+(aq)<---

FinnZ [79.3K]

Answer: $Kc=1.24*10^5^2$

Explanation: For the given reaction:

$Kc=\frac{[Zn^+^2]}{[Ag^+]^2}$

Concentrations of the ions are not given so we need to think about another way to calculate Kc.

We can calculate the free energy change using the standard cell potential as:

$\Delta G^0=-nFE^0_c_e_l_l$

$E^0_c_e_l_l$ can be calculated using standard reduction potentials.

Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.

$E^0_c_e_l_l$ = $E^0_c_a_t_h_o_d_e+E^0_a_n_o_d_e$

Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.

$E^0_c_e_l_l$ = 0.78 V - (-0.76 V)

$E^0_c_e_l_l$ = 0.78 V + 0.76 V

$E^0_c_e_l_l$ = 1.54 V

Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .

$\Delta G^0=-(2*96485*1.54)$

$\Delta G^0$ = -297173.8 J

Now we can calculate Kc using the formula:

$\Delta G^0=-RTlnKc$

T = 25+273 = 298 K

R = $8.314JK^-^1mol^-^1$

--297173.8 = -(8.314*298)lnKc

297173.8 = 2477.572*lnKc

$lnKc=\frac{297173.8}{2477.572}$

lnKc = 119.946

$Kc=e^1^1^9^.^9^4^6$

$Kc=1.24*10^5^2$

5 0

3 years ago

In the formula for calcium nitrate below, how many atoms of nitrogen are represented?

yaroslaw [1]

Answer:

2HNO3+Ca(OH)2 ---> Ca(NO3)2+2H2O

you put the two on the H2O to make it 2H2O and A two on the HNO3 to make it 2HN03 To make it balanced

Explanation:

hope i could help every one but i'm only one person i've answered 20 (now 21) questions today my goal is to get 4 brainlyest maybe 8 if i'm lucky

8 0

1 year ago

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