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puteri [66]
1 year ago
14

A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant

figures?
Chemistry
2 answers:
VLD [36.1K]1 year ago
6 0

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

                           \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

Tatiana [17]1 year ago
3 0

The volume of the gas at {\text{100 }}^\circ{\text{C}} is \boxed{{\text{17}}{\text{.5 L}}}

Further explanation:

Charles’s law:

Charles’s work showed that at constant pressure, the volume-temperature relationship for a fixed amount of gas is linear. In other words, Charles’s law can be stated that at constant pressure, the volume occupied by a fixed amount of a gas is directly proportional to its absolute temperature (Kelvin). This relationship is known as <em>Charles’s law</em>.

The mathematical representation of Charles’s law is,

{\mathbf{V}}\propto{\mathbf{T}}                                             [P and n are constant]

Here,

 V is volume occupied by the fixed quantity of gas.

 T is the temperature of a gas.

 P is the pressure of a gas.

 n denotes the number of moles of gas.

The relationship can also be expressed as,

\frac{{\text{V}}}{{\text{T}}}={\text{constant}}                               [P and n are constant]

Or it can also be expressed as follows:

\frac{{{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}=\frac{{{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}                          …… (1)

Here,

{{\text{V}}_1} is the initial volume of gas.

{{\text{V}}_2} is the final volume of gas.

{{\text{T}}_1} is the initial temperature of the gas.

{{\text{T}}_2} is the final temperature of the gas.

Rearrange equation (1) to calculate {{\text{V}}_2}.

{{\text{V}}_{\text{2}}}=\frac{{{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}                           …… (2)

The value of {{\text{V}}_1} is 3.50 L.

The value of {{\text{T}}_1} is {\text{20 }}^\circ{\text{C}}.

The value of {{\text{T}}_2} is {\text{100 }}^\circ{\text{C}}.

Substitute these values in equation (2).

\begin{gathered}{{\text{V}}_{\text{2}}}=\frac{{\left({{\text{3}}{\text{.50 L}}}\right)\left({{\text{100 }}^\circ{\text{C}}}\right)}}{{\left({{\text{20 }}^\circ{\text{C}}}\right)}}\\={\mathbf{17}}{\mathbf{.5 L}}\\\end{gathered}

The volume of gas occupied at{\mathbf{100 ^\circ C}}is 17.5 L.

Learn more:

1. Law of conservation of matter states: brainly.com/question/2190120

2. Calculation of volume of g as: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas of equation

Keywords: Charles’s law, volume, temperature, pressure, volume temperature relationship, absolute temperature, constant pressure, relationship, V directly proportional to T, ideal gas, ideal gas equation number of moles, moles, P, n, V, T, volume of gas, 17.5 L, Kelvin, 3.50 L,

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Answer:

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Explanation:

Los parámetros dados son;

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Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

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5 0
1 year ago
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
1 year ago
6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets
Nata [24]

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: V_{1} = 0.20 mL,         T_{1} = 27^{o}C

V_{2} = ?,                    T_{2} = 177^{o}C

Formula used to calculate new volume is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL

Thus, we can conclude that the new volume of cake is 1.31 mL.

5 0
1 year ago
How much heat is required to heat 9.61g of ethanol (CH3CH2OH) from 24.10C to 67.30C?
Licemer1 [7]

Answer:

a.)  

To warm the liquid from 35°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - 35) = 4154 J

To vaporize the liquid at 78°C:

(38.56 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 35.154 kJ

Total:

4.154 kJ + 35.154 kJ = 39.3 kJ

b.)  

To warm the solid from -155°C to -114°C:

(0.97 J/g-K) x (42.0 g) x (-114°C - (-155°C)) = 1670 J

To melt the solid at -114°C:

(5.02 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 4.5766 kJ  

To warm the liquid from -114°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - (-114)) = 18547 J

To vaporize the liquid at 78°C:

35.154 kJ  (as in part a.)

Total:

1.670 kJ + 4.5766 kJ + 18.547 kJ + 35.154 kJ = 59.9 kJ

Explanation:

8 0
2 years ago
A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in
Fittoniya [83]

The molarity (M) of the acid in the lake water is 0.00001M .

<u>Explanation:</u>

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

M_1V_1=M_2V_2

Here;

M_1,M_2  are molarity of acid in the lake water and base solution respectively.V_1,V_2  are volume of sample in the lake water and base solution respectively.

Given values are as follows:

M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml

Putting these values in above equation :

M_1V_1=M_2V_2

M_1(1000)=(0.100)(0.100)

M_1=\frac{(0.100)(0.100)}{1000}

M_1=0.00001M

Therefore, the molarity (M) of the acid in the lake water is 0.00001M .

5 0
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