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GaryK [48]
1 year ago
5

As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a den

sity of 0.993 g/mL. The difference in heights of the liquid in the two sides of the manometer is 32.3mm when the atmospheric pressure is 765 mm Hg. Given that the density of mercury is 13.6 g/mL, the pressure of the enclosed gas is ________ atm. Group of answer choices
Chemistry
1 answer:
STALIN [3.7K]1 year ago
5 0

Answer:

1.01atm is the pressure of the gas

Explanation:

The difference in heights in the two sides is because of the difference in  pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:

32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg

As atmospheric pressure is 765mm Hg and assuming the gas has more pressure than the atmospheric pressure (There is no illustration), the pressure of the gas is:

765mm Hg + 2.36mm Hg = 767.36 mmHg

In atm:

767.36 mmHg * (1atm / 760 mmHg) =

1.01atm is the pressure of the gas

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A flask contains methane, chlorine and carbon monoxide gases. the partial pressures of each are 0.215 atm, 0.066 atm, and 0.826
hjlf

Answer;

The total pressure is 1.107 atm.

Explanation;

The total pressure is the sum of the pressures of the three gases in the flask

Pressure (total) = 0.215 atm + 0.066 atm + 0.826 atm = 1.107 atm

= 1.107 atm.

8 0
2 years ago
Read 2 more answers
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.

B)

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.

C)

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
Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be
Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

                       In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

                             Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

                    HCOOH (formic acid)    →     HCOO⁻ (formate)  +  H⁺

                H₃CCOOH (acetic acid)     →      H₃CCOO⁻ (acetate)  +  H⁺

Formal Charges:

                           Formal charges are calculated using following formula,

          F.C  =  [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

                    F.C  =  [6] - [6 + 2/2]

                    F.C  =  [6] - [6 + 1]

                    F.C  =  6 - 7

                    F.C  =  -1

For Sodium:

                    F.C  =  [1] - [0 + 0/2]

                    F.C  =  [1] - [0]

                    F.C  =  1 - 0

                    F.C  =  +1

5 0
2 years ago
An electrically neutral atom of gallium has 31 electrons and 39 neutrons. What is the mass number for an atom of gallium?. A. 31
hammer [34]
If the atom is neutral (meaning, not charged) the number of electron is equal to the number of protons. The mass number of an atom is the sum of the number of proton and the number of neutrons. From the given above, the mass number of gallium is 31 + 39. The answer is letter D. 70.
4 0
2 years ago
Read 2 more answers
The hybrid orbital set used by the central atom in h2co is ________. the hybrid orbital set used by the central atom in h2co is
Feliz [49]
The Lewis structure for H₂CO is shown in the attached picture. The central atom is the carbon. However, I'm not sure which bond you're referring to. There can be two answers. The two C-H bonds are sp³ hybridized because it is a single bond. The C=O bond is sp² hybridized because it is a double bond.

5 0
2 years ago
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