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2 years ago

How many total bond are in the Lewis structure for HSIN?

Chemistry

1 answer:

enyata [817]2 years ago

6 0

Answer:

The number of bonds for a neutral atom is equal to the number of electrons in the full valence shell (2 or 8 electrons) minus the number of valence electrons. This method works because each covalent bond that an atom forms adds another electron to an atoms valence shell without changing its charge.

Explanation:

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How many mg does an 830 kg sample contain?

Aleksandr-060686 [28]

Answer:

A sample of 830 kg would contain 830000000 mg

Explanation:

3 0

1 year ago

What is a common factor associated with most severe storms?

Volgvan

A common factor is low pressure system.

4 0

1 year ago

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Boron - 10 emits alpha particles and cesium - 137 emits beta particles. Write balanced nuclear

anzhelika [568]

Answer:

B10 5N +5P= Li6 3N +3P

Cs 137 82N+55P = I 133 80N + 53P

Explanation:

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2 years ago

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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

1 year ago

Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2. What is

timama [110]

Answer:

4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel

Explanation:

The reaction given in the question is

$2Al + 3 Br_2$ ⇒ $2AlBr_3$

According to the stoichiometric coefficients of the reaction, 2 moles of $Al$ requires 3 moles of $Br_2$ so in this reaction, $Br_2$ is a limiting reagent. So we will consider that $Al$ is in excess.

Now,

Since 3 moles of $Br_2$ requires 2 moles of $Al$

So, for 6 moles of $Br_2$ the moles of $Al$ required = $\frac{2}{3} \times 6$ = 4 moles.

Moles of $Al$ remaining after the completion of reaction = 5 - 4 = 1 mole.

Again,

Since 3 moles of $Br_2$ produces 2 moles of $AlBr_3$

So, moles of $AlBr_3$ produced by 6 moles of $Br_2$ = $\frac{2}{3} \times 6$ = 4 moles.

Therefore, after the completion of reaction, 4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel.

5 0

2 years ago