A sample of chlorine gas has a volume of 0.30 L at 273 K and 1 atm pressure. What temperature (in ∘C) would be required to increase the volume to 1.0 L ?

Answers

Answer 1

Answer:

[tex]T_2=637\°C[/tex]

Explanation:

Hello there!

In this case, considering this problem as pressure constant, since the change is exhibited in temperature and volume only, it is possible for us to use the Charles' law as shown below:

[tex]\frac{V_2}{T_2}=\frac{V_1}{T_1}[/tex]

Thus, by solving for the final temperature, we obtain:

[tex]T_2=\frac{T_1V_2}{V_1}\\\\T_2=\frac{273K*1L}{0.30L}\\\\T_2=910K-273\\\\T_2=637\°C[/tex]

Best regards!


Related Questions

Helllllp me please
Part D ends with neon Express your answer as an integer. 20

Answers

Answer:

attach file please

Question attached above

Answers

The correct answer is option B or number two electrons are consumers

A gas has a pressure of 699.0 mm Hg at 40.0 0C. This temperature is
K.

What is the temperature at 760.0 mm Hg? K (round to 1 decimal place) or

degrees Celsius (round to one decimal place.

Answers

Answer:

initial temp: 313.2 Kfinal temp: 340.5 Kin Celsius: 67.3 °C

Explanation:

Given a volume and quantity of gas at 699.0 mm Hg pressure has a temperature of 40.0 °C, you want to know the temperature at 760.0 mm Hg pressure.

Pressure and Temperature

Pressure and absolute temperature of a gas are proportional for a given quantity and volume.

Initial temperature

The temperature of 40.0 °C is an absolute temperature of ...

  40.0 K + 273.15 K = 313.15 K ≈ 313.2 K

New temperature

The temperature after increasing the pressure can be found from ...

  T/313.15 = (760.0 mm)/(699.0 mm)

  T = 313.15 · 760/699 ≈ 340.5 . . . . Kelvin

In Celsius, that is ...

  340.5 K -273.2 = 67.3 °C

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34 g of O2 are reacted with excess Cs, causing a production of 199 g of Cs2O. What is the percent yield of this
reaction?
2 Cs + O2 ---> Cs2O

Answers

Answer:

33.23 %

Explanation:

4 Cs + O₂ → 2Cs₂O

First we convert 34 g of O₂ into moles, using its molar mass:

34 g O₂ ÷ 32 g/mol = 1.0625 mol O₂

Then we convert O₂ moles into Cs₂O moles, using the stoichiometric coefficients of the balanced reaction:

1.0625 mol O₂ * [tex]\frac{2molCs_2O}{1molO_2}[/tex] = 2.125 mol Cs₂O

Now we convert 2.125 moles of Cs₂O into grams, using its molar mass:

2.125 mol Cs₂O * 281.81 g/mol = 598.85 g Cs₂O

598.85 g is the theoretical yield. Finally we proceed to calculate the percent yield:

199 / 598.85 * 100% = 33.23 %


A gas occupies a volume of 2.4 L at 0.14 ATM. What volume will the gas occupy at 0.84 ATM?

Answers

Answer:

0.4 L

Explanation:

Calculate by using Boyle's Law P₁V₁=P₂V₂

(0.14atm)(2.4L) = (0.84atm)(V₂)

0.336 atmL = (0.84atm)(V₂)

V₂ = 0.336 atmL/0.84atm

V₂ = 0.4 L

arrange the following group of atoms in order of increasing atomic size:B,Al,Ga​

Answers

Answer:

Al,Ga,B

Explanation:

Now since i helped you can you help me with this plz

Matteo took 5 math quizzes. The mean of the 5 quizzes was 8.2. Here are four of his quiz scores 7, 7, 8, 10. What is the 5th quiz score? Show work.

Boron ( B )

Aluminium ( Al )

Gallium ( Ga )

Write a balanced nuclear equation for the beta decay of 234/90Th

Answers

The balanced nuclear equation for the beta decay of 234/90Th can be represented as follows: ^234/90Th --> ^234/91Pa + e^0/-1β

In this equation, the nucleus of thorium-234 (234/90Th) undergoes beta decay. During beta decay, a neutron within the nucleus is converted into a proton, resulting in the emission of an electron (beta particle). As a result, the thorium-234 nucleus is transformed into protactinium-234 (234/91Pa) by gaining one proton.

The beta particle emitted during the decay process is represented as e^0/-1β, where the superscript 0 denotes that the electron has no charge (neutral), and the subscript -1 indicates that the electron carries a negative charge of -1.

It is important to note that in a nuclear equation, the total atomic mass and atomic number on both sides of the equation must be equal to maintain a balanced equation and conserve mass and charge.

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g At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H 2 (g) Br 2 (g) 2HBr (g) A mixture of 0.682 mol of H 2 and 0.440 mol of Br 2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.516 mol of H 2 present. At equilibrium, there are ________ mol of Br 2 present in the reaction vessel.

Answers

Answer: At equilibrium , there are 0.274 moles of [tex]Br_2[/tex]

Explanation:

Moles of  [tex]H_2[/tex] = 0.682 mole

Moles of  [tex]Br_2[/tex] = 0.440 mole

Volume of solution = 2.00 L

Initial concentration of [tex]H_2[/tex] = [tex]\frac{0.682}{2.00}=0.341 M[/tex]

Initial concentration of [tex]Br_2[/tex] =  [tex]\frac{0.440}{2.00}=0.220 M[/tex]

Equilibrium concentration of [tex]H_2[/tex] = [tex]\frac{0.516}{2.00}=0.258 M[/tex]

The given balanced equilibrium reaction is,

             [tex]H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)[/tex]

Initial conc.   0.341 M    0.220 M        0 M    

At eqm. conc.   (0.341-x) M   (0.220-x) M   (2x) M

Given : (0.341-x) M = 0.258 M

x= 0.083 M

Thus equilibrium concentartion of [tex]Br_2[/tex] = (0.220-0.083) M = 0.137 M

Thus moles of [tex]Br_2[/tex] at equilibrium = [tex]0.137M\times 2.00L=0.274mol[/tex]

At equilibrium , there are 0.274 moles of [tex]Br_2[/tex]

At 40°C how many grams of sugar would dissolve in 100 grams of water

Answers

Answer:

I think is C 240 yup yup

Explanation:

At 40°C how many grams of sugar would dissolve in 100 grams of water?

prop-1-yne + 2HBr/H2O2 = A;
A + 2H2O = B;
B + K2CO3(aq) = C;
C + heat = D;
D + HBr = E.
find the compounds A, B, C, D and E

Answers

Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3/s, and the discharge from the treatment plant is 37 m3/s. Upstream from the outfall, the background concentration is 0.69 mg/L. The maximum allowable concentration in the river is 1.1 mg/L. What is the maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant

Answers

Answer:

[tex]2.54\ \text{mg/L}[/tex]

Explanation:

C = Allowable concentration = 1.1 mg/L

[tex]Q_1[/tex] = Flow rate of river = [tex]130\ \text{m}^/\text{s}[/tex]

[tex]Q_2[/tex] = Discharge from plant = [tex]37\ \text{m}^3/\text{s}[/tex]

[tex]C_1[/tex] = Background concentration = 0.69 mg/L

[tex]C_2[/tex] = Maximum concentration that of the pollutant

The concentration of the mixture will be

[tex]C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}[/tex]

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is [tex]2.54\ \text{mg/L}[/tex].

Bases are typically found in

Answers

Answer:

they are typically found in

1)soap

2) detergent

3) baking soda etc..

All chemical reactions use reactants in a specific proportion or stoichiometry to regulate the amount of products produced. The __________ reactant is consumed completely during the reaction but some of the __________ reactant is left over.
a. limiting, excess
b. excess, limiting
c. proportional, ideal
d. chemical, stoichiometric

Answers

Answer: The answer is a. limiting, excess. I took the test.

Explanation:




Please help meeee, I will brainliest. and
Put it into your own words, if u use from internet.

Answers

Answer:

energy is used to break bonds in reactants and energy is released when new bonds form in products.The law of conservation of energy states that matter cannot be created or destroyed. Whether a chemical reaction absorbs or releases energy there is no overall change in the amount of energy during the reaction.

Explanation:

Sorry if im wrong

Atoms with 5 or more valence electrons tend to form __________________ ions, called _________.

Answers

Answer:

positively-charged  cations

Explanation:

hope this helps! :)

what is an example of an electrolyte solution?

Answers

Answer:

nacl with water

they are capable of conducting electricity

What is it called when a gas turns into a liquid

Answers

Answer:

condensation

Explanation:

The process of a gas becoming a liquid is called condensation.

☁️ Answer ☁️

Condensation

"Condensation is the name of the process through which gas turns into a liquid. Three common states of matter exist: solids, liquids and gases. Matter, however, can change its state with the addition or subtraction of energy. The energy is usually in the form of heat. "

Hope it helps.

Have a nice day noona/hyung.

A sample of O2 occupies 75 L at 1 atm. If the volume of the
sample doubles, what is the new pressure of Oz?

Answers

Answer:

the  new pressure of Oz is 0.5 atm

Explanation:

The computation of the new pressure of Oz is shown below:

Given that

Original pressure of O2 (P1) = 1 atm

And, the Original volume of oxygen O2 (V1) = 75L

Now

New pressure of O2 (P2) = ?

And, the New volume of O2 (V2) = 2V1  i.e

= (2 × 75L)

= 150L

Here we applied the Boyle's law formula  

P1V1 = P2V2

1 atm × 75L = P2 × 150L

75 atm × L = 150L ×P2

Now Divide both sides by 150L to get P2

75 atm × L ÷ 150L = 150L × P2 ÷ 150L

0.5 atm = P2

Hence, the  new pressure of Oz is 0.5 atm

?CH4 → ?C6H6 + ?H2

?C2H6 + ?O2 → ?CO2 + ?H2O

I need to know the number that goes in front of the equations

Answers

Answer:

You basically want to balance these equations.

(6)CH4 → (1)C6H6 + (9)H2

(2)C2H6 + (7)O2 → (4)CO2 + (6)H2O

NOTE: DO NOT PUT BRACKETS IN YOUR ANSWER, I HAVE PUT HERE FOR THE ANSWERS INSTEAD OF QUESTION MARKS.

6CH4->1C6H6 + 9H2

2C2H6 + 702-> 4CO2 + 6H2O

Which statement is accurate about variations?

Organisms with variations that give an advantage are more likely to survive

Organisms with variations are less likely to pass these variations to their offspring

In all cases vanations become more common in a population

In all cases variations affect how likely it is that an organism reproduces

Answers

Answer:

A,C,D

Explanation:

Variations effect the likelyhood of survival for a organism, it can give off an advantage or disadvantage.

Brainliest? pls

Question List
Question 4 of 8
Total Points: 0 out of 80
Calculate the volume that will be occupied by
240 mL of hydrogen, measured at 360 mm Hg, when
the pressure is changed to 1080 mm Hg.

Answers

Answer:

80 mL

Explanation:

Step 1: Given data

Initial pressure of hydrogen (P₁): 360 mmHgInitial volume of hydrogen (V₁): 240 mLFinal pressure of hydrogen (P₂): 1080 mmHgFinal volume of hydrogen (V₂): ?

Step 2: Calculate the final volume of hydrogen

If we assume ideal behavior and constant temperature, we can calculate the final volume occupied by hydrogen using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 360 mmHg × 240 mL/1080 mmHg = 80 mL

what is the pH if the H+ is 5.7 x 10^-10 M ?

Answers

Answer:

9.24

Explanation:

-logofH

Which factor contributes the most to the rates of diffusion and effusion between two gases in a mixture?
a)The size of the particles.
b)The molar mass of the particles.
c)The interactions between the particles.
d)The relative volume of the particles to each other.

Answers

Answer:

b)The molar mass of the particles.

Explanation:

The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham's law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases.

Gallium (, ) is a metalloid obtained from its salts during the smelting of ores of other elements, like Zinc. has broad applicability in the electronics industry. It is also used as a safe replacement for mercury in thermometers as it melts at and has a heat of fusion of . What is the entropy change of of gallium in as it melts when placed on a surface at

Answers

The given question is incomplete. The complete question is:

Gallium (Ga, 69.723 g/mol) is a metalloid obtained from its salts during the smelting of ores of other elements, like Zinc. has broad applicability in the electronics industry. It is also used as a safe replacement for mercury in thermometers as it melts at 29.8 °C and has a heat of fusion of 5.59 kJ/mol. What is the entropy change of 22 g of gallium in J/K as it melts when placed on a surface at 29.8°C?

Answer: The entropy change of of gallium in as it melts when placed on a surface at  is 5.81 J/K

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to melt 1 mole of Ga = 5.59 kJ

Mass of Ga given = 22 gram

Heat required to melt 69.723 g of Ga = 5.59 kJ

Thus Heat required to melt 22 g of Ga = [tex]\frac{5.59kJ}{69.723}\times 22=1.76kJ[/tex]

Temperature = [tex]29.8^0C=(273+29.8)kJ=302.8kJ[/tex]

Now entropy change = [tex]\frac{\text {heat of fusion}}{\text {temperature in K}}=\frac{1.76kJ}{302.8K}=0.00581kJ/mol=5.81J/K[/tex]

Thus the entropy change of of gallium in as it melts when placed on a surface at [tex]29.8^0C[/tex] is 5.81 J/K

HELP QUICK!!!!!!
How many valence electrons do the atoms in Group 13 have?

Answers

Answer:

The answer is 3 valance electrons.

Explanation:

Hope this helped Mark BRAINLIEST!!!

Answer:

3 valence electrons

Explanation:

The atoms in group 13 have three valence electrons.

Heat is most closely related to which kind of energy?

Answers

Answer:

I'm very sure it's thermal energy.

Explanation:

Answer:

thermal energy is the answer

How many polyatomic ions (s) below contain a total of four atoms?

hydroxide ion
nitrate ion
ammonium ion
sulfite ion


Group of answer choices

two

four

three

one

Answers

Answer:

I think the answer is two

Considering change of state descriptions, select the name of the change of state for when energy is absorbed. Group of answer choices sublimation freezing condensation deposition

Answers

Answer: Sublimation

Explanation:

Sublimation : It is a process in which a solid directly changes to gaseous phase by providing heat.

Freezing is a process in which a liquid changes into solid phase when allowed to cool.

Condensation is a process in which a gas changes into liquid phase when allowed to cool.

 Deposition is a  process in which a gas deposits as solid phase when allowed to cool.

Thus change of state where energy is absorbed is sublimation.

Convert 8.19 x 10-14 moles of gold to atoms.

Answers

Answer:

4.932018 × 10¹⁰ atoms

Explanation:

Recall that:

1 mole = 6.022 × 10²³ atoms

To change 8.19× 10⁻¹⁴ moles to atoms;

Then, we have:

= (8.19× 10⁻¹⁴× 6.022 × 10²³ atoms)

= 4.932018 × 10¹⁰ atoms

Which of the following is an example of synthesis?


A) HgO + Cl2-->HgCl + O2

B) Na + Br2 --> NaBr

C) KClO3 --> KCl + O2

D) Cl2 + NaBr --> NaCl + Br2

Answers

Answer:

B

Explanation:

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