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2 years ago

Write the complete nuclear symbol for a nucleus of oxygen with 8 protons & 10 neutrons?

Chemistry

1 answer:

Sav [38]2 years ago

4 0

Answer: The complete nuclear symbol will be $_8^{18}\textrm {O}$

Explanation:

The nuclear symbol is written as: $_Z^A\textrm {X}$ where Z= atomic number , A = mass number X = symbol of element

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

For an electrically neutral atom,

Atomic number = Number of electrons = Number of protons = 8

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

Mass number = 8 + 10 = 18

Thus complete symbol will be $_8^{18}\textrm {O}$

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

2 years ago

Is a pencil a compound or an element and is a pencil heterogeneous or homogeneous

masya89 [10]

Answer:

Pencil is an element and a heterogeneous mixture

Explanation:

6 0

2 years ago

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Non-conjugated β, γ-unsaturated ketones, such as 3-cyclohexenone, as in base-catalyzed equilibrium with their α,β-unsaturated is

uranmaximum [27]

Answer:

See the attached file for the structure.

Explanation:

Find attached for the explanation

7 0

2 years ago

7.) Regarding molecular formula and empirical formula for a particular compound... (A) the subscripts in the empirical formula m

S_A_V [24]

Answer:

Explanation:

i think the anwer would be that its elliptical orbit and the tilt an its axis.

3 0

2 years ago

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Calculate the whole-number ratio of staggered to eclipsed conformers that are present at room temperature. Use the Actual G˚ bar

natka813 [3]

Answer:

The ratio of staggered to eclipsed conformers is 134

Explanation:

It is possible to determine the ratio of staggered to eclipsed conformers of a reactant, using the equilibrium:

Staggered ⇄ Eclipsed

Keq = [Eclipsed] / [Staggered]

That means Keq is equal to the ratio we need to find:

Using:

G˚= -RTln Keq

<em>Where G° = -12133.6J/mol</em>

<em>R is gas constant: 8.314J/molK</em>

<em>T is absolute temperature: 298K</em>

<em />

-12133.6J/mol= -8.314J/molK*298K ln Keq

4.8974 = ln Keq

134 = Keq = [Eclipsed] / [Staggered]

<h3>The ratio of staggered to eclipsed conformers is 134</h3>

3 0

2 years ago