If you observe a Full Moon on January 12th, on what date would you observe the next New Moon?

A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 2 atm
T1= 50 C= 323 K (being 0 C= 273 K)
P2= 3.2 atm
T2= ?

Replacing:

$\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}$

Solving:

$T2*\frac{2 atm}{323 K} =3.2 atm$

$T2=3.2 atm*\frac{323 K}{2 atm}$

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

5 0

3 years ago

ANSWER ASAP!

Rainbow [258]

Answer:

All around you there are chemical reactions taking place. Green plants are photosynthesising, car engines are relying on the reaction between petrol and air and your body is performing many complex reactions. In this chapter we will look at two common types of reactions that can occur in the world around you and in the chemistry laboratory. These two types of reactions are acid-base reactions and redox reactions.

Explanation:

7 0

3 years ago

Read 2 more answers

Which agent would most likely

fredd [130]

Answer:

Wind

Explanation:

Most conifers and about 12% of the world's flowering plants are wind-pollinated. Wind pollinated plants include grasses and their cultivated cousins, the cereal crops, many trees, the infamous allergenic ragweeds, and others. All release billions of pollen grains into the air so that a lucky few will hit their targets.

7 0

3 years ago

A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of

mina [271]

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.125

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.125=e^{-0.00440\times t}$

t = 473 year

8 0

3 years ago

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not

frutty [35]

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

A + B → C

the rate law can be expressed as:

Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

Rate₁ = k[A]₁ᵃ[B]ᵇ → eq. 1

Rate₂ = k[A]₂ᵃ[B]ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

(2) = (2)ᵃ
taking log of both sides
log (2) = a Log (2)
0.693 = a * 0.693
a =1

∴ order of reaction with respect to A is first (=1) → (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

Rate₁ = k[A]ᵃ[B]₁ᵇ → eq. 1

Rate₂ = k[A]ᵃ[B]₂ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

(1) = (2)ᵇ
taking log of both sides
log (1) = b Log (2)
0 = 0.693 * b
b = 0

∴ order of reaction with respect to B is zero → (2)

So, from 1 and 2 the right choice is rate = k[A]¹[B]⁰= k[A]

6 0

3 years ago