A mixture of h2 and water vapor is present in a closed vessel at 20.00°c. the total pressure of the system is 755.0 mmhg. the partial pressure of the water vapor 17.50 mmhg. what is the partial pressure of h2? type in your answer using the correct number of significant figures. mmhg

Answer: 139.2 g

Explanation:

2.9 mol KClO3 (3 mol O2 / 2 mol KClO3) (32 g / 1 mol O2) = 139.2 grams :)

The solute in the scenario where a chemist adds 36.0 g of sodium chloride to 200 g of water at STP and all of the sodium chloride is sodium chloride.

A solute is a substance that forms a solution with another substance called solvent in a process called solubility.

According to this question, a chemist adds 36.0 g of sodium chloride to 200 g of water at STP and all of the sodium chloride dissolves in the water.

This shows that the sodium chloride is the solute added to the water (solvent).

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Explanation:

Ethanoic acid is a weak acid and dissociates:

C

H

3

C

O

O

H

⇌

C

H

3

C

O

O

−

+

H

+

For which

K

a

=

[

C

H

3

C

O

O

−

]

[

H

+

]

[

C

H

3

C

O

O

H

]

=

1.8

×

10

−

5

Please remember that these are equilibrium concentrations .

p

K

a

=

−

log

K

a

=

−

log

[

1.8

×

10

−

5

]

=

4.74

n

=

c

×

v

∴

n

C

H

3

C

O

O

H

=

0.20

×

200.0

1000

=

0.04

Taking into account the definition of calorimetry, the temperature change of A will be one sixth of the temperature change of B.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, for object A:

QA = cA× mA× (ΔT)A

and for object B:

QB = cB× mB× (ΔT)B

You know:

Object A has double the specific heat of object B. ⇒ cA= 2× cB

Object A has triple the mass of object B. ⇒ mA= 3× mB

If the same amount of heat is applied to both objects ⇒ QA= QB

Then:

cA× mA× (ΔT)A= QB = cB× mB× (ΔT)B

2× cB × 3× mB× (ΔT)A=  cB× mB× (ΔT)B

Solving:

6× cB × mB× (ΔT)A= cB× mB× (ΔT)B

(ΔT)A= (cB× mB× (ΔT)B) ÷ (6× cB × mB)

(ΔT)A=[tex]\frac{1}{6}[/tex] (ΔT)B

Finally, the temperature change of A will be one sixth of the temperature change of B.

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Answer: 6.7 kPa

Explanation:

Pressure Total = P(partial 1) + P(partial 2) + P(partial 3) →

94.5 kPa = 65.4 kPa + 22.4 kPa + P(partial 3) →

6.7 = P(partial 3) :)

According to the Dalton's law of partial pressures the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas in the mixture.

So, P[tex]_o[/tex] = P[tex]_1[/tex]+ P[tex]_2[/tex]+ P[tex]_3[/tex]

[tex]\longrightarrow[/tex]94.5 = 65.4+ 22.4 + P[tex]_3[/tex]

[tex]\longrightarrow[/tex]94.5 = 87.8+ P[tex]_3[/tex]

[tex]\longrightarrow[/tex] P[tex]_3[/tex] = 94.5 - 87.8

[tex]\longrightarrow[/tex]P[tex]_3[/tex] = 6.7 kpa

Therefore, partial pressure of the 3rd gas of the mixture is 6.7 kpa .

The extraction of metal usually makes use of the method of precipitation in which a reagent is added which is able to precipitate the required metal in its pure form without impurities.

The process of extraction has to do with the removal of a substance from another. It is usually the way that pure substances are obtained in chemistry. The question is incomplete but I will try to give a general idea.

The extraction of metal usually makes use of the method of precipitation in which a reagent is added which is able to precipitate the required metal in its pure form without impurities. We can not make numerica;l calculatiosn as the values are missing.

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For an unknown using the retention times 1. 6 min for ch4, 12. 8 min for octane, 15. 4 min for the unknown, and 19. 6 min for nonane, the kovats retention index is mathematically given as

I = 807.857

Generally, the equation for the Kovats index(i)  is mathematically given as

[tex]I= \frac{100[n + (N - n) x (Log tr (unknown) - logtr (n))}{ logtr(N) - logtr(n))]}[/tex]

Therefore

[tex]I= \frac{100[ 1+8*(log(14.7) -log (1.5)}{log (19.8) - log (1.5)) 0]}[/tex]

I = 100[ 1+8*(0.991/1.12)]

I = 807.857

In conclusion, Kovats retention index

I = 807.857

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Kovats retention index is a system that converts the retention time into the independent constants of the system. The Kovats retention index for the unknown is 807.857.

The Kovats retention index (I) is the system used in analytical techniques like gas chromatography for the conversion of the retention time.  It is given mathematically as:

[tex]\rm I = \rm \dfrac{100[n+(N-n)x(Logtr(unknown)-logtr(n))}{Logtr(N)-Logtr(n)]}[/tex]

Substituting values and solving further,

[tex]\begin{aligned} \rm I &= \rm \dfrac{100[1+8 \times (Log(14.7) - log (1.5))}{(log (19.8) - log (1.5))0}\\\\&= 100 [ 1 + 8 (\dfrac{0.991}{1.12})]\\\\&= 807.857\end{aligned}[/tex]

Therefore, the Kovats retention index for an unknown is 807.857.

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#SPJ4

The final pressure or the air pressure on the top of the mountain is 0.818.

Air pressure is the pressure exerted by air in the atmosphere of the earth.

By the Gas law

[tex]\dfrac{P_1V_1}{ T1} =\dfrac{ P_2V_2 }{T_2}[/tex]

Given, the pressure of the valley floor is 1 atm. = P1

The temperature is 25° C = T1

The volume of the bag is 0.985 L. = V1

The temperature at the top of the mountain is ° C = T2

Now, the volume of the bag is 1. 030 L = V2

To find P2

Put the values in the formula

[tex]\dfrac{1 atm \times0.985 L}{ 25} =\dfrac{ P_2 \times 1. 030 L }{22}\\\\0.03832 = 0.0468 \times P_2\\\\P_2= 0.818[/tex]

Thus, the final pressure is 0.818.

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The new volume of the gas in the balloon when the pressure is doubled is determined as 11.2 L.

The new volume of the balloon is determined by applying general gas equation as shown below

P₁V₁/T₁ = P₂V₂/T₂

Where;

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\\\V_2 = \frac{P_1V_1T_2}{T_1P_2} \\\\V_2 = \frac{1 \times 22.4 \times 273}{273 \times 2} \\\\V_2 = 11.2 \ L[/tex]

Thus, the new volume of the gas in the balloon when the pressure is doubled is determined as 11.2 L.

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75% dominant and 25% resistive

When Lead reacted with nitric acid, lead(II) nitrate, nitrogen oxide and water are produced.

What products are produced when lead reacts with nitric acid?

Lead(II) nitrate, nitrogen oxide and water are the products which are produced by reacting lead and nitric acid with each other. The nitrogen attach with the lead molecule as well as oxygen which produces Lead(II) nitrate and nitrogen oxide.

So we can conclude that when Lead reacted with nitric acid, lead(II) nitrate, nitrogen oxide and water are produced.

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Answer:

9   I believe

Explanation:

(a) When more gas is added, the number of moles of the gas increases, and  the total pressure of gas inside same vessel will increase.

(b) When the conatiner's volume is increased, the total pressure of gas inside the vessel will decrease.

The ideal gas law of general gas equation, can be used to describe the behavior of an ideal gas.

The ideal gas law is given as;

PV = nRT

where;

When more gas is added, the number of moles of the gas increases, and  the total pressure of gas inside same vessel will increase.

When the conatiner's volume is increased, the total pressure of gas inside the vessel will decrease.

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Answer:

All of the signals pass through the thalamus

Explanation:

C. Walking on wet floors shows that Friction is undesirable

Since K3 is greater than 1, the reaction is thermodynamically favorable.

Data;

This states that the heat change in a given chemical reaction is the same irrespective of the number of process in which the reaction is effected.

Using Hess Law;

[tex]rxn_3 = rxn_1 * 2rxn_2\\k_3 = k_1 * k_2^2\\k_3 = 2*10^-^6 * (1*10^3)^2\\k_3 = 2[/tex]

substituting this using Gibbs law,

[tex]\delta G^0_3 = -RT\ln K_3\\\delta G^0_3 < 0[/tex]

Since K3 is greater than 1, the reaction is thermodynamically favorable.

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Answer:

The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

at t=0  cM              0             0

at eqm                    

So dissociation constant will be:

Give c= 4.4 M and  = ?

Putting in the values we get:

Also

Thus pH of a 4.4 M  solution is 4.3

Answer:

These clouds extend through the lower and mid-layers of the troposphere bringing rain to the surface below.

Explanation:

The bases of clouds in the middle level of the troposphere, given the prefix “alto,”

Answer:

An organism's role within an ecosystem depends on how it obtains its food. Plants and animals obtain their food in very different ways, so they have very different roles in an ecosystem. The way in which an organism obtains food also affects its interactions with other organisms in the ecosystem.

Explanation:

Answer:

B . Water molecules can collide with other particles more than neon atoms can.

✥ So, in 1.25 moles of NO2, you would have 1.25 moles of N and (2 x 1.25) 2.50 moles of O.

0.1 mole = Avogadro's number of particles (approx. 6.022x10^23 particles/mol)

So, #oxygen atoms = 2.50 mol x 6.022x10^23 atoms/mol

6.022x10^23 atoms/molOr, approximately 1.51x10^24 oxygen atoms.

Hope this answer helps you ✨

0 J

Work = Force * Displacement

Since displacement is 0 m (object isn't moving), work would also be 0.

Answer:

84.0 g

Balanced equation: 4Fe + 3O2 --> 2Fe2O3

Find moles of oxygen:

Find mass of O2

Answer:

7. H

8. F

9. G.

10. D

11. A

12. E

13. B

14. C

Answer:

Arsenic (33As) has 33 known isotopes and at least 10 isomers.

The buret should be conditioned before loading it with the titrant.

This is defined as laboratory apparatus which is used in quantitative chemical analysis to measure the volume of a liquid or a gas.

It should be conditioned with the titrant solution to prevent dilution from occurring as a result of cleaning it with water.

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Answer:

Polyester

Explanation:

1)

methyl

CH3 (mol)

CH3- (struc)

2)

ethyl

C2H5

CH3-CH2-

3)

propyl

C3H7

CH3-CH2-CH2-

4)

butyl

C4H9

CH3-CH2-CH2-CH2-

5)

pentyl

C5H11

CH3-CH2-CH2-CH2-CH2-

6)

hexyl

C6H13

CH3-CH2-CH2-CH2-CH2-CH2-

7)

heptyl

C7H15

CH3-CH2-CH2-CH2-CH2-CH2-CH2-

8)

octyl

C8H17

i will stop writing the condensed formula ust add another CH2

9)

nonyl

C9H19

10)

decyl

C10H21

11)

Undecyl

C11H23

12)

dodecyl

C12H25