When an oxygen atom forms an ion, it gains two electrons. What is the electrical charge of the oxygen ion? A.
+2

B.
-1

C.
+1

D.
-2

Answers

Answer 1

Answer: D. 2- youre adding two negative charges to a neutral element

Answer 2

Answer:

The electrical charge of the oxygen ion is -2.

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Predict the organic product formed when phosgene reacts with an excess of methanol.
When converting an acid salt dissolved in water to its acid form, you are instructed to adjust the pH well into the acidic range (pH = ~2). Why don’t you just take the pH to neutral? You do the same for adjusting the pH of a solution containing an amine salt taking it well into the basic range rather than neutral. Why are you instructed to do this?

Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2*10^-4

Answers

This problem uses the relationship between Kb and the the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:

Kb = KaKb
1.00 x 10^-14 = 7.2 x 10^-4(x)
x = 1.39 x 10^-11

We now need to calculate the [OH¯] using the Kb expression:

1.39 x 10^-11 = x^2 / (0.30 - x)

The denominator can be neglected. Thus, x is 3.73 x 10^-6.

pOH = -log 3.73 x 10^-6 = 5.43
pH = 14-5.43 = 8.57

Answer:

pH=8.32

Explanation:

The relevant equilibrium for this problem is

F⁻ + H₂O ↔ HF + OH⁻

With a constant Kb of

Kb= $([HF][OH^(-)])/([F^(-)])$

Kb= $(x*x)/(0.30-x)$

To calculate the value of Kb we use the formula Kw=Ka*Kb, where Kw is the ionization constant of water, 1 * 10⁻¹⁴.

1 * 10⁻¹⁴ = 7.2*10⁻⁴ * Kb

Kb = 1.4 * 10⁻¹¹

So now we have

1.4 * 10⁻¹¹= $(x*x)/(0.30-x)$

We make the assumption that x<<<0.30 M, so we can rewrite the equation of Kb as:

1.4 * 10⁻¹¹= $(x*x)/(0.30)$

$4.2*10^(-12)=x^(2) \nx=2.05*10^(-6)$

So [OH⁻]=2.05*10⁻⁶

pOH=5.68

pH = 14 - pOH

pH=8.32

20 Which term is defined as a measure of the disorder of a system?(1) heat (3) kinetic energy
(2) entropy (4) activation energy

Answers

Entropy is the term defined as a measure of the disorder of a system. This occurs mainly when the thermal energy is not available for the mechanical things.

Heat is a measure of energy in a medium
Kinetic energy is a measure of energy in an object that's resulting in movement
Activation energy is the amount of energy required to start a chemical reaction
Entropy is a measure of disorder in a system
The answer is Entropy

How many carbon atoms are there in a diamond (pure carbon) with a mass of 77 mg

Answers

Answer:

3.86 x 10^24 carbon atoms in a 77 gram diamond

Explanation:

A key tool for chemistry and physics is the concept of the mole. The definition of a mole is actually quite simple. One mole is 6.02 x 10^23 of something. Anything. One can say " I have 1 mole of paperclips(PPC) in my closet.". That simply says that there are 6.02 x 10^23 paperclips(PPC) in your closet. It does suggest, however, that you may be demented. One mole of PPC, placed end-to-end, would stretch a LONG way.

Average PPC Length = 4.45 cm

! mole of PPC would stretch (6.02 x 10^23)*(4.45 cm) = 2.68 x 10^24 cm

That is 2.68 x 10^19 km

The distance to the moon is 3.84x10^5 km. So 1 mole of PPC will stretch to the moon and back 6.97 x 10^13 times.

This mental diversion does not answer the question, but it does explain what comes next for calculating the number of carbon atoms in 77 mg of diamond. The brilliant connection that allows this calculation is how 1 mole (Avogadro's Number) ties in with the atomic masses of the elements. The atomic mass of an element, as shown on a periodic table, is also the mass of that element that contain Avogadro's Number of atoms. The atomic mass, written in AMU (atomic mass units) can ALSO be written as the molar mass, in units of grams/mole. That is, the number of grams of that element that would amount to 6.02 x 10^23 (1 mole) of tat element's atoms.

Carbon, with an atomic mass of 12, can be said to have a Molar Mass of 12 grams/mole.

Use the units to guide the calculation needed to find the number of carbon atoms (in this case divide the grams carbon by the molar mass):

(77 grams Carbon)/(12 grams/mole Carbon)

The grams unit cancels and the moles unit moves to the top, leaving a number that is simply moles of carbon.

(77 grams Carbon)/(12 grams/mole Carbon) = 6.42 moles carbon

By definition, 1 mole equals 6.02 x 10^23 carbon atoms.

(6.42 moles carbon)*((6.02 x 10^23 carbon atoms)/mole)

Moles cancel, leaving carbon atoms

= 3.86 x 10^24 carbon atoms in a 77 gram diamond

How much energy is passed to the next trophic level in the food chain?

Answers

Answer:

10%

Explanation:

Out of the original layer, only ten percent is transferred to the next

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 3.50 g of butane?

Answers

First convert grams to molesusing molar mass of butane that is 58.1 g

3.50g C4H10 x (1 molC4H10)/(58.1g C4H10) = 0.06024 mol C4H10

Now convert moles to molecules by using Avogadro’s number

0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 molC4H10) = 3.627x10^22 molecules C4H10

And there are 4 carbon atoms in 1 molecule of butane, so usethe following ratio:

3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 moleculeC4H10)
= 1.45 x 10^23 atoms of carbon are present

$\boxed{{\text{1}}{\text{.45}} * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ atoms}}}$ of carbon is present in 3.50 g of butane.

Further Explanation:

Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.

Avogadro’s number is the number of units that are present in one mole of the substance. Its value is equal to ${\text{6}}{\text{.022}}*{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$ per mole of substance. These units can be electrons, atoms, molecules or ions.

The formula to calculate the moles of butane is as follows:

${\text{Moles of butane}}=\frac{{{\text{Given mass of butane}}}}{{{\text{Molar mass of butane}}}}$ ......(1)

The given mass of butane is 3.50 g.

The molar mass of butane is 58.12 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Moles of butane}}&=\left({{\text{3}}{\text{.50 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.12 g}}}}}\right)\n&={\text{0}}{\text{.060220234 mol}}\n&\approx {\text{0}}{\text{.0602 mol}}\n\end{aligned}$

The molecules of butane are calculated as follows:

${\text{Molecules of butane}}=\left({{\text{Moles of butane}}}\right)\left( {{\text{Avogadro's Number}}}\right)$ ......(2)

The moles of butane is 0.060220234 mol.

The value of Avogadro’s number is ${\text{6}}{\text{.022}}*{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ .

Substitute these values in equation (2).

$\begin{aligned}{\text{Molecules of butane}}{\mathbf&{ = }}\left({0.060220234{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} *{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}}\right)\n&={\text{3}}{\text{.62646}}*{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}\n&\approx {\text{3}}{\text{.626}}*{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{molecules}}\n\end{aligned}$

One molecule of butane has four carbon atoms. So the number of carbon atoms in ${\text{3}}{\text{.62646}}* {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}$ of butane is calculated as follows:

${\text{Atoms of carbon}}=\left( {{\text{Molecules of butane}}}\right)\left( {\frac{{{\text{4 carbon atoms}}}}{{{\text{1 butane molecule}}}}}\right)$

.......(3)

Substitute ${\text{3}}{\text{.62646}}*{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}$ for the molecules of butane in equation (3).

$\begin{aligned}{\text{Atoms of carbon}}&=\left( {{\text{3}}{\text{.62646}}* {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}}\right)\left({\frac{{{\text{4 carbonatoms}}}}{{{\text{1 molecule}}}}} \right)\n&= 1.45 * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ atoms}}\n\end{aligned}$

Therefore the number of atoms of carbon is ${\mathbf{1}}{\mathbf{.45 * 1}}{{\mathbf{0}}^{{\mathbf{23}}}}{\mathbf{ atoms}}$ .

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603

2. Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: mole, atoms of carbon, molecules of butane, carbon-12, 12 g, Avogadro’s number, butane, moles of butane, electrons, molecules, atoms.

Which radioisotope is used for diagnosing thyroid disorders?(1) U-238 (3) I-131
(2) Pb-206 (4) Co-60

Answers

The correct answer is option 3. The radioisotope used for diagnosing thyroid disorders is I-131. This isotope is used due to its low expense compared to other radioisotopes. I- 31 is a beta and gamma emitter used for the ablations of thyroid tumors.

Final answer:

I-131 (Iodine-131) is the radioisotope used for diagnosing thyroid disorders. It is selectively taken up by the thyroid gland and its emissions allow for imaging of the gland.

Explanation:

The radioisotope used for diagnosing thyroid disorders is I-131 (Iodine-131). This radioisotope is used in medical procedures due to its unique characteristics. When consumed, it is selectively taken up by the thyroid gland. I-131 emits beta particles and gamma radiation which allows for imaging of the thyroid gland. This in turn helps health professionals diagnose if the thyroid is functioning normally or if disorders such as hyperthyroidism or tumors are present.