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3 years ago

WHAT WILL HAPPEN TO THE RESISTANCE OF CONDUCTOR IF :

Chemistry

1 answer:

Debora [2.8K]3 years ago

3 0

Answer:

1. The length of the wire is halved.

the resistance of a conductor is directly proportional to it's length (L) as R - L. Thus doubling it's length will double it's resistance, while halving it's length would halve the resistance. Also the resistance of a conductor is inversely proportional to it's cross sectional area.

2. The area of cross-section of the conductor in increased.

On increasing the area of cross-section, resistance decreases. Thus is because resistance is inversely proportional to area.

3. The temperature of the conductor is increased.

With increasing temperature, the resistance of the wire increases as collisions within the wire increases and "slow" the flow of current... Since conductors typically display an increased resistivity with temperature increase, they have a positive temperature coefficient.

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2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

what is the ratio of the rate of effusion of helium (atomic mass 4.00 amu) to that of oxygen gas (molecular mass 32.0 amu)?

nignag [31]

Answer:

3 : 1

Explanation:

Let the rate of He be R1

Molar Mass of He (M1) = 4g/mol

Let the rate of O2 be R2

Molar Mass of O2 (M2) = 32g/mol

Recall:

R1/R2 = √(M2/M1)

R1/R2 = √(32/4)

R1/R2 = √8

R1/R2 = 3

The ratio of rate of effusion of Helium to oxygen is 3 : 1

8 0

3 years ago

What is a conversion factor?

Zarrin [17]

Answer: an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

Explanation:

5 0

3 years ago

Read 2 more answers

consider the following equilibrium:h2co3(aq) h2o(l) h3o (aq) hco3-1(aq).what is the correct equilibrium expression?

MAVERICK [17]

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:

H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]

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3 years ago

Ga3+ and Br1- is what formula?

Sladkaya [172]

$\text{GaBr}_3$

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3 years ago