Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg,what is the atmospheric pressure

Answers

Answer 1

Answer:

760 mmHg

Explanation:

Step 1: Given data

Partial pressure of nitrogen (pN₂): 592 mmHgPartial pressure of oxygen (pO₂): 160 mmHgPartial pressure of argon (pAr): 7 mmHgPartial pressure of the trace gas (pt): 1 mmHg

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg


Related Questions

what is the difference between acidic and basic protein​

Answers

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10 genes.

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10

Explanation:

Carbon-14 has a half-life of 5720 years and this is a fast-order reaction. If a piece of wood has converted 75 % of the carbon-14, then how old is it?

Answers

Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

Drag each image to the correct location on the model. Each image can be used more than once. Apply the rules and principles of electron configuration to draw the orbital diagram of aluminum. Use the periodic table to help you.

Answers

Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

what is the meaning of the word tetraquark?​

Answers

Answer:

A tetraquark in physics is an exotic meson composed of four valence quarks.

Explanation:

It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.

Hope it helps.

Identify a homogeneous catalyst:

a. SO2 over vanadium (V) oxide
b. H2SO4 with concentrated HCl
c. Pd in H2 gas
d. N2 and H2 catalyzed by Fe
e. Pt with methane

Answers

Answer:

b, H2SO4 with HCl, as they are both liquid acids

Limiting reagent problem. How many grams of H2O is produced from 40.0 g N2O4 and 25.0 g N2H4. N2O4 (l) + 2 N2H4 (l) → 3 N2 (g) + 4 H2O(g)

Answers

Answer:

28.13 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol

Mass of N2O4 from the balanced equation = 1 x 92 = 92g

Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol

Mass of N2H4 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72 g

Summary:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4.

Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.

From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.

Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.

Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.

In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.

The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:

From the balanced equation above,

64 g of N2H4 reacted to produce 72 g of H2O.

Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.

Therefore, 28.13 g of H2O were obtained from the reaction.

After the reaction between sodium borohydride and the ketone is complete, the reaction mixture is treated with water and H2SO4 to produce the desired alcohol. Explain the reaction by clearly indicating the source of the hydrogen atom that ends up on the oxygen

Answers

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

What is the molar mass of a protein if a solution of 0.020 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 ∘ C

Answers

Answer:

26.5 kD  

Explanation:

Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.56 - ( given ). This is only one part of the information we are given / can conclude in this case ....

i = van’t Hoff factor = 1 for a protein molecule,

R = gas constant = 62.36 L torr / K-mol,

T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K

( Known initially ) ∏ = osmotic pressure = 0.56 torr

..... besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.

_____

Substitute derived / known values to solve for M ( moles / liter ) -

∏ = iMRT

⇒ 0.56 = ( 1 )( M )( 62.36 )( 298 )

⇒ 0.56 = M( 18583.28 )

⇒ M = 0.56 / 18583.28 ≈ 0.00003013461 ....

_____

We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -

M = mol / l

⇒ 0.00003013461 = 0.020 / 25 mL ( 0.025 L ),

0.020 / 0.025 = 0.8 g / L

⇒ 0.8 g = 0.00003013461 moles,

molar mass = 0.8 g / 0.00003013461 moles = 26,548 g / mol = 26.5 kD  

If enough experimental data supports a hypothesis, then it

Answers

Answer:

Then the hypothesis is proved and becomes a theory.

If not, then another hypothesis should be proposed and tested.

Night vision glasses detect
energy emitted from cooling objects?
ultraviolet
infrared
X-ray

Answers

Answer:

I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.

Explanation:

16. A metal element and a non-metal element are brought near each other and allowed to react. What's the most likely type of compound
that will form between these two elements?
A. lonic and covalent
B. lonic
C. Covalent
D. Neither, metals and non-metals don't react.​

Answers

B) ionic. Generally, metals and non-metals form ionic bonds that are stronger than covalent bonds due to the higher electronegativity difference.

Answer:

B) Ionic

Explanation:

When a solution is diluted with water, the ratio of the initial to final
volumes of solution is equal to the ratio of final to initial molarities
Select one:
True
-​

Answers

Hello!!

The correct answer for this problem would be TRUE.

Explanation: it is true that when a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities.

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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A student mixes wants to prepare 24.1 mmol of benzamide from benzoyl chloride and NH4OH. If the student uses excess 15 M NH4OH, how many mL of Benzoyl chloride must be used

Answers

Answer:

2.81mL

Explanation:

Based on the reaction:

C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl

Benzoyl chloride + ammonia → Benzamide

1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.

Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.

As molar mass of benzoyl chloride is 141g/mol, mg you require are:

mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.

to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:

3.3981g × (1mL / 1.21g) =

2.81mL

A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be

Answers

Answer:

25.99mL is the volume internal volume of the flask

Explanation:

To complete the question:

The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask

The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.

To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:

Mass water = Mass filled flask - Mass of clean flask

Mass water = 60.167g - 34.232g

Mass water = 25.935g of water.

To convert this mass to volume:

25.935g × (1mL / 0.997992g) =

25.99mL is the volume internal volume of the flask

AB2AB2 has a molar solubility of 3.72×10−4 M3.72×10−4 M. What is the value of the solubility product constant for AB2AB2? Express your answer numerically.

Answers

Answer:

Ksp = 2.06x10⁻¹⁰

Explanation:

For AB₂. solubility product constant, Ksp, is written as follows:

AB₂(s) ⇄ A²⁺ + 2Br⁻

Ksp = [A²⁺] [Br⁻]²

Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:

AB₂(s) ⇄ A²⁺ + 2Br⁻

AB₂(s) ⇄ X + 2X

where X are moles that are soluble (Molar solubility)

Ksp = [X] [2X]²

Ksp = 4X³

As molar solubility of the salt is 3.72x10⁻⁴M:

Ksp = 4X³

Ksp = 4(3.72x10⁻⁴)³

Ksp = 2.06x10⁻¹⁰

How many mL of calcium hydroxide are required to neutralize 25.0 mL of 0.50 M
nitric acid?

Answers

Answer:

6.5 mL

Explanation:

Step 1: Write the balanced reaction

Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O

Step 2: Calculate the reacting moles of nitric acid

25.0 mL of 0.50 M  nitric acid react.

[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]

Step 3: Calculate the reacting moles of calcium hydroxide

The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol

Step 4: Calculate the volume of calcium hydroxide

To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.

[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]

cetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.

Answers

Answer:

0.60 mol

Explanation:

There is some info missing. I think this is the original question.

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.

Step 1: Given data

Moles of water required: 1.5 mol

Step 2: Write the balanced equation

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water

The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol

below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (g) ΔH1°= 449.2 kJ R2 O3 (g) + Cl (g) → O2 (g) + ClO (g) ΔH2° = -126 kJ R3 ClO (g) + O (g) → O2 (g) + Cl (g) ΔH3°= -268 kJ

Answers

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

ΔH = -338.8kJ

The table below shows the electronegativity values of various elements on the periodic table. Electronegativities A partial periodic table. Which pair of atoms would form a covalent bond ? calcium (Ca) and bromine (Br) rubidium (Rb) and sulfur (S) cesium (Cs) and nitrogen (N) oxygen (O) and chlorine (Cl)

Answers

Answer:

Oxygen and Chlorine

Explanation:

Covalent bonds involve the sharing of electrons between nonmetals.

Answer:

oxygen (O) and chlorine (Cl)

Explanation:

cuz i said so

The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g of latic acid is used to make a solution with a volume of 1.00l what is the concentration of hydronium ion in the solution

Answers

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling point. Finally, explain how you came to that conclusion.

Answers

Answer:

Ethane would have a higher boiling point.

Explanation:

In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.

Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.

I hope it helps!

The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.

We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.

We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.

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A sample of neon gas at a pressure of 0.609 atm and a temperature of 25.0 °C, occupies a volume of 19.9 liters. If the gas is compressed at constant temperature to a
volume of 12.7 liters, the pressure of the gas sample will be
atm.​

Answers

Answer:

The pressure of the gas sample will be  0.954 atm.​

Explanation:

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

To determine the change in pressure or volume during a transformation at constant temperature, the following is true:

P1 · V1 = P2 · V2

That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.

In this case:

P1= 0.609 atmV1= 19.9 LP2=?V2= 12.7 L

Replacing:

0.609 atm* 19.9 L= P2* 12.7 L

Solving:

[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]

P2= 0.954 atm

The pressure of the gas sample will be  0.954 atm.​

Need help finding major products

Answers

Answer:

Explanation:

RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻

C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅

C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻

In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺

CH₃CH₂CH₂CH₂CH₂⁺ ⇒  CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

Hence option D is correct .

b )

In the second case carbocation produced is

CH₃CH₂CH₂CH⁺CH₃

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

The product formed is same as in case of first

Option B is correct

A saturated solution was formed when 5.16×10−2 L of argon, at a pressure of 1.0 atm and temperature of 25 ∘C, was dissolved in 1.0 L of water.
Calculate the Henry's law constant for argon. it must be im M/atm

Answers

Answer:

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

C = k*P

where C is the solubility, P the partial pressure and k is the Henry constant.

So, being the concentration [tex]C=\frac{ngas}{V}[/tex]  

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]

In this case:

P=PAr= 1 atmV=VAr= 5.16*10⁻² LR=0.082 [tex]\frac{atm*L}{mol*K}[/tex]T=25 °C=298 °K

Then:

[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]

Solving:

n= 2.11 *10⁻³ moles

So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]

Using Henry's Law and being C=CAr and P =PAr:

2.11*10⁻³ M= k* 1 atm

Solving:

[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]

You get:

[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.

What is Henry's law?

Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.

C∝P

C = kP, where

k = Henry's constant

P = partial pressure of gas

C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V

n = no. of moles

V = volume

And moles of the gas will be calculated by using the ideal gas equation as:

PV = nRT

n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles

And Concentration in liquid will be:

C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M

Now we put all these values in the first equation to calculate the value of k as:

k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm

Hence required value of k is 2.1 × 10⁻³M/atm.

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Why was it important to establish the Clean Air Act?

Answers

Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.

Explanation:

Answer:

Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.

Calculate the mass of feso4 that would be produced by 0.5mole of Fe

Answers

Answer:76 grams

Explanation:

Fe+H₂SO₄-->FeSO₄+H₂

For one mole of Fe we get 1 mole of feso4, therefore for 0.5 moles of Fe we get 0.5 moles of feso4.

The molar mass of feso4 is AFe+AS+4AO(A is atomic mass)

56+32+4*16=152grams/mole

Now, we need to multiply the number of moles by the molar mass to get the mass that reacts

152*0.5=76 grams

Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the ions that would be present in solution. Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation. Express your answer as a chemical equation including phases.

Answers

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

An aqueous solution containing 5.06 g of lead(II) nitrate is added to an aqueous solution containing 6.03 g of potassium chloride.Enter the balanced chemical equation for this reaction. Be sure to include all physical states.balanced chemical equation:What is the limiting reactant?lead(II) nitratepotassium chlorideThe percent yield for the reaction is 82.9% . How many grams of the precipitate are formed?precipitate formed:gHow many grams of the excess reactant remain?excess reactant remaining:

Answers

Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl

11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KCl solution? Are these results what you would expect? Why?

Answers

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
P2O5 (s) + H2O (l) =H3PO4 (aq)
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.

Answers

Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

2. We'll begin by writing the balanced equation for the reaction. This is given below:

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

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