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Bond [772]
2 years ago
10

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu

me was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?
Chemistry
1 answer:
katrin2010 [14]2 years ago
8 0

Answer:

T2=276K

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

(V1/T1)=(V2/T2)

T2=( V2*T1)/V1

T2=(322*298)/348

T2=276K

Hence, the temperature of the freezer is 276 K

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How many grams of KClO3 are needed to produce of 4.26 moles of O2? 2 KClO3 2 KCl + 3 O2 a. 348 g b. 136 g c. 174 g d. 522 g e. 7
JulsSmile [24]
Ok so this is what we know :

2KClO3 -> 2KCl + 3O2         (Always check if equation is balanced - in this                                               case it is)
                              4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.

R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
                      <span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
                       348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).

8 0
1 year ago
Read 2 more answers
Describe the many different forms of energy involved with stretching and releasing a rubber band. What other processes are simil
Nadya [2.5K]

Answer:

Conversion of kinetic energy to potential energy (chemo mechanical energy)

In the state of rest, the rubber is a tangled mass of long chained cross-linked polymer that due to their disorderliness are in a state of increased entropy. By pulling on the polymer, the applied kinetic energy stretches the polymer into straight chains, giving them order and reducing their entropy. The stretched rubber then has energy stored in the form of chemo mechanical energy which is a form of potential energy

Conversion of the stored potential energy in the stretched to kinetic energy

By remaining in a stretched condition, the rubber is in a state of high potential energy, when the force holding the rubber in place is removed, due to the laws of thermodynamics, the polymers in the rubber curls back to their state of "random" tangled mass releasing the stored potential energy in the process and doing work such as moving items placed in the rubber's path of motion such as an object that has weight, w then takes up the kinetic energy 1/2×m×v² which can can result in the flight of the object.

Explanation:

5 0
1 year ago
Read 2 more answers
Consider the general reversible reaction. Lower A upper A plus lower B upper B double-headed arrow lower C upper C plus Lower d
dimulka [17.4K]

Answer: K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. It is represented by the symbol K_{eq}

The balanced chemical reaction is:

aA+bB\rightleftharpoons cC+dD     

The expression for K_{eq} is written as:

K_{eq}=\frac{[C]^c\times [D]^d}{[A]^a\times [B]^b}

Thus the correct option is K e q equals StartFraction StartBracket upper C EndBracket superscript lower c StartBracket upper D EndBracket superscript lower D over StartBracket upper A EndBracket superscript lower a StartBracket upper B EndBracket superscript lower b EndFraction.

3 0
2 years ago
Small quantites of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc.
ira [324]

<u>Answer:</u> The mass or zinc reacted is 0.624 grams.

<u>Explanation:</u>

We are given:

Total pressure = 1.032 atm

Vapor pressure of water = 32 torr = 0.042 atm    (Conversion factor:  1 atm = 760 torr)

To calculate partial pressure of hydrogen gas, we use the equation:

p_{H_2}=p_T-p_{H_2O}\\\\p_{H_2}=1.032-0.042=0.99atm

To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure of hydrogen gas = 0.99 atm

V = Volume of hydrogen gas = 240. mL = 0.240 L    (Conversion factor: 1 L = 1000 mL)

T = Temperature of hydrogen gas = 30^oC=[30+273]K=303K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

0.99atm\times 0.240L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303K\\n=\frac{0.99\times 0.240}{0.0821\times 303}=9.55\times 10^{-3}mol

The chemical equation for the reaction of zinc and hydrochloric acid follows:

Zn+2HCl\rightarrow ZnCl_2+H_2

By Stoichiometry of the reaction:

1 mole of hydrogen gas is produced from 1 mole of zinc metal

So, 9.55\times 10^{-3}mol of hydrogen gas is produced from = \frac{1}{1}\times 9.55\times 10^{-3}=9.55\times 10^{-3}mol of zinc metal

To calculate the mass of zinc metal, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of zinc = 65.38 g/mol

Moles of zinc = 9.55\times 10^{-3} moles

Putting values in above equation, we get:

9.55\times 10^{-3}mol=\frac{\text{Mass of zinc}}{65.38g/mol}\\\\\text{Mass of zinc}=(9.55\times 10^{-3}mol\times 65.38g/mol)=0.624g

Hence, the mass or zinc reacted is 0.624 grams.

4 0
2 years ago
A sample of krypton has a volume of 6.00 L, and the pressure is 0.960 atm. If the final temperature is 55.0°C, the final volume
MissTica

193.38 K was the initial temperature of the krypton.

Explanation:

Data given:

Initial volume of the krypton gas = 6 litres

initial pressure of the krypton gas = 0.960 atm

initial temperature of the krypton gas = ?

final volume of the krypton gas  = 7.70 litres

final pressure of the Krypton gas  = 1.25 atm

final temperature of the krypton gas = 55 degrees or 273.25+55 = 323.15 K

Applying the  Combined Gas Laws:

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Rearranging the equation:

T1  = \frac{P1V1T2}{P2V2}

Putting the value in the equation:

T1 = \frac{0.960 X 6 X 323.15}{1.25 X 7.70}

T1 = 193.38 K

Initial temperature of the krypton gas is 193.78 K

5 0
2 years ago
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