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4 years ago

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Which of the following is not a type of radiation? A. trans-radiation B. inhaled radiation C. terrestrial radiation D. cosmogeni

c radiation

2 answers:

Sindrei [870]4 years ago

5 0

I would say the answer is A.

elixir [45]4 years ago

5 0

I think its D. because that makes the most sense if you look at all of the answers

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What is the most common acid and base on earth?

nata0808 [166]

Answer:

hydrochloride acid

Explanation:

hope this helps :D

5 0

3 years ago

Determine the pH of (a) a 0.40 M CH3CO2H solution, (b) a solution that is 0.40 M CH3CO2H and 0.20 M NaCH3CO2

elena-s [515]

Answer:

a) pH = 2.573

b) pH = 4.347

Explanation:

a) weak acid: CH3COOH

CH3COOH + H2O ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = [H3O+][CH3COO-] / [CH3COOH]

∴ <em>C</em> CH3COOH = 0.40 M

mass balance:

⇒ 0.40 M = [CH3COO-] + [CH3COOH].........(1)

charge balance:

⇒ [H3O+] = [CH3COO-].........(2)

(2) in (1):

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ Ka = 1.8 E-5 = [H3O+]² / ( 0.40 - [H3O+] )

⇒ [H3O+]² = 7.2 E-6 - 1.8 E-5[H3O+]

⇒ [H3O+]² + 1.8 E-5[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 2.6743 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.573

b) balanced reations:

CH3COONa + H2O → Na+ + CH3COO-
CH3COOH + H2O ↔ CH3COO- + H3O+

∴ <em>C</em> CH3COOH = 0.40 M

∴ <em>C</em> CH3COONa = 0.20 M

mass balanced:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [CH3COO-] + [CH3COOH]

⇒ 0.60 = [CH3COO-] + [CH3COOH]......(1)

charge balanced:

⇒ [H3O+] + [Na+] = [CH3COO-]

∴ [Na+] = 0.20 M

⇒ [H3O+] + 0.20 M = [CH3COO-]........(2)

(2) in (1):

⇒ 0.60 M = ( [H3O+] + 0.20 ) + [CH3COOH]

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ 1.8 E-5 = ([H3O+])([H3O+] + 0.20) / (0.40 - [H3O+])

⇒ 7.2 E-6 - 1.8 E-5[H3O+] = [H3O+]² + 0.20[H3O+]

⇒ [H3O+]² + 0.20[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 4.499 E-5 M

⇒ pH = 4.347

7 0

4 years ago

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.

<u>Explanation:</u>

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:

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

<u>For $_{30}^{64}\textrm{Zn}$ isotope:</u>

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

<u>For $_{30}^{66}\textrm{Zn}$ isotope:</u>

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

<u>For $_{30}^{67}\textrm{Zn}$ isotope:</u>

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

<u>For $_{30}^{68}\textrm{Zn}$ isotope:</u>

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

<u>For $_{30}^{70}\textrm{Zn}$ isotope:</u>

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0

3 years ago

The free energy for the oxidation of glucose to CO2 and water is -686 kcal/mol, and the free energy for the reduction of NAD+ to

vitfil [10]

Answer:

Most of free energy available from oxidation of the glucose remains in pyruvate.

Explanation:

The overall reaction of the process glycolysis is:

Glucose + 2 NAD⁺ + 2 ADP + 2 Pi ⇒ 2 Pyruvate + 2 NADH + 2 H⁺ + 2ATP

Glucose is oxidized to give 2 molecules of pyruvate and 2 molecules of NADH and ATP (Energy currency).

<u>Though the free energy of oxidation of glucose is high but only 2 NADH is formed because the most of the free energy that is being released from the oxidation of glucose remains in the pyruvate which is produced in the reaction and thus only 2 molecules are formed.</u>

6 0

3 years ago

What is polar satellites????

katrin [286]

Answer:

polar orbit is one in which a satellite passes above or nearly above both poles of the body being orbited (usually a planet such as the Earth, but possibly another body such as the Moon or Sun) on each revolution. It has an inclination of about 60 - 90 degrees to the body's equator.[1] A satellite in a polar orbit will pass over the equator at a different longitude on each of its orbits.

Launching satellites into polar orbit requires a larger launch vehicle to launch a given payload to a given altitude than for a near-equatorial orbit at the same altitude, due to the fact that much less of the Earth's rotational velocity can be taken advantage of to achieve orbit. Depending on the location of the launch site and the inclination of the polar orbit, the launch vehicle may lose up to 460 m/s of Delta-v, approximately 5% of the Delta-v required to attain Low Earth orbit. Polar orbits are a subtype of Low Earth orbits with altitudes between 200 and 1,000 kilometers.[1]

Explanation:

8 0

3 years ago