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4 years ago

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Which of the following elements, essential to life, is a trace element? Which of the following elements, essential to life, is a

trace element? phosphorus carbon calcium iodine

1 answer:

Vikentia [17]4 years ago

7 0

Answer:

The answer to your question is Iodine

Explanation:

Elements can be classified as:

Basic elements of life: Carbon, Hydrogen, Oxygen and, Nitrogen these elements are in great quantity in the body.

Quantity elements are elements in a medium quantity in the body like Sodium, Magnesium, Potassium, Calcium, Phosphorus, Sulfur and Chlorine.

Trace elements are elements in a low quantity example: Selenium, Iodine, Zinc, Copper, Cobalt, Iron and, Manganese.

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Ethyl ethanoate undergoes following reaction [on the picture]

vichka [17]

<u>Answer:</u> The concentration of ethyl ethanoate at equilibrium is $0.653mol/dm^3$

<u>Explanation:</u>

Given values:

Equilibrium concentration of ethanol = $0.42mol/dm^3$

Equilibrium concentration of ethanoic acid = $0.42mol/dm^3$

$K_c=0.27$

The given chemical equation follows:

$CH_3COOC_2H_5+H_2O\rightleftharpoons C_2H_5OH+CH_3COOH$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[C_2H_5OH][CH_3COOH]}{[CH_3COOC_2H_5]}$

Putting values in above expression, we get:

$0.27=\frac{0.42\times 0.42}{[CH_3COOC_2H_5]}$

$[CH_3COOC_2H_5]=\frac{0.42\times 0.42}{0.27}=0.653mol/dm^3$

Hence, the concentration of ethyl ethanoate at equilibrium is $0.653mol/dm^3$

5 0

3 years ago

A scientist measures the standard enthalpy change for the following reaction to be 81.1 kJ :2CO2(g) + 5 H2(g)C2H2(g) + 4 H2O(g)B

posledela

<u>Answer:</u> The enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.Z

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

For the given chemical reaction:

$2CO_2(g)+5H_2(g)\rightarrow C_2H_2(g)+4H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ$

Putting values in above equation, we get:

$81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol$

Hence, the enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.

8 0

3 years ago

The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c

Ivahew [28]

Explanation:

(a) The given data is as follows.

B = $2.180 \times 10^{-18} J$

Z = 4 for Be

Now, for the first excited state $n_{f}$ = 2; and $n_{i} = \infinity$ if it is ionized.

Therefore, ionization energy will be calculated as follows.

I.E = $\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})$

= $8.72 \times 10^{-18} J$

Converting this energy into kJ/mol as follows.

$8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol$

= 5249 kJ/mol

Therefore, the ionization energy of the $Be^{3+}$ ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

$\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}$

$\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}$

$\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}$

= $303.7 \times 10^{-10} m$

or, = $303.7^{o}A$

Therefore, wavelength of light given off from the $Be^{3+}$ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels $303.7^{o}A$ .

5 0

3 years ago

Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to b

DIA [1.3K]

<h3>Answer:</h3>

Longest wavelength = 343.7 nm

<h3>Solution and Explanation:</h3>

In this question we need to first use the concept of energy of a photon.

Energy of a photon, E, is given by the formula, E = hf, where h is the plank's constant, f is the frequency.

But since, f is given by dividing speed, c, by wavelength, λ, then;

E = hc/λ

We are given 348 kJ/mol required to break carbon-carbon bonds.

We know that; 1 mole of bonds = 6.022 × 10^23 bonds.

We are required to find the longest wavelength with enough energy to break the C-C bonds.

This can be worked out in simple steps:

Step 1: Energy required to break one bond (kJ/bond)

1 mole of bonds = 6.022 × 10^23 bonds.

Therefore;

348 kJ = 6.022 × 10^23 bonds.

Thus;

1 bond = 348 kJ ÷ 6.022 × 10^23 bonds.

= 5.778 x 10^-22 kJ

But; 1000 joules = 1 kJ

Hence; energy per bond = 5.778 x 10^-19 Joules

Step 2: Energy per photon

Breaking one bond requires energy equivalent to energy of a photon.

Therefore;

1 photon = 5.778 x 10^-19 Joules

= 5.778 x 10^-19 J/photon

Step 3: Calculating the wavelength

From the equation of energy of a photon;

E = hc/λ

h is the plank's constant = 6.626 × 10^-34 J/s

c is the speed of light in vacuum = 2.9998 × 10^8 m/s

E is the energy of a photon = 5.778 x 10^-19 Joules

Therefore, making λ (wavelength) the subject;

$wavelength = \frac{hc}{E}$

$= \frac{(6.626 . 10^{-34})(92.9998.10^8) }{(5.778 .10^{-19} )}$

$= 3.437. 10^{-7} m$

= 3.437 x 10^-7 m

But; 1 nm = 10^-9 m

Thus;

wavelength = 343.7 nm

Therefore, the longest wavelength of the radiation will be 343.7 nm

5 0

3 years ago

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tatyana61 [14]

Pure metals possess few important physical and metallic properties, such as melting point, boiling point, density, specific gravity, high malleability, ductility, and heat and electrical conductivity. These properties can be modified and enhanced by alloying it with some other metal or nonmetal, according to the need.

Alloys are made to:

Enhance the hardness of a metal: An alloy is harder than its components. Pure metals are generally soft. The hardness of a metal can be enhanced by alloying it with another metal or nonmetal.

Lower the melting point: Pure metals have a high melting point. The melting point lowers when pure metals are alloyed with other metals or nonmetals. This makes the metals easily fusible. This property is utilized to make useful alloys called solders.

Enhance tensile strength: Alloy formation increases the tensile strength of the parent metal.

Enhance corrosion resistance: Alloys are more resistant to corrosion than pure metals. Metals in pure form are chemically reactive and can be easily corroded by the surrounding atmospheric gases and moisture. Alloying a metal increases the inertness of the metal, which, in turn, increases corrosion resistance.

Modify color: The color of pure metal can be modified by alloying it with other metals or nonmetals containing suitable color pigments.

Provide better castability: One of the most essential requirements of getting good castings is the expansion of the metal on solidification. Pure molten metals undergo contraction on solidification. Metals need to be alloyed to obtain good castings because alloys

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3 years ago