All categories

3 years ago

What is a real life example of Charles’ law?

1 answer:

max2010maxim [7]3 years ago

6 0

Leaving a basketball out in the cold weather. When a basketball if left in a cold garage or outside during the cold months, it loses its air inside (or volume).

You might be interested in

I need someone to help with the bonus.

siniylev [52]

Answer:

268.3 m/sec

Explanation:

7 0

3 years ago

What are prevailing winds?

Tems11 [23]

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart

7 0

3 years ago

why is hydrogen a nonmetal,usually placed with group 1 elements in the periodic table even though it does not show a metallic pr

Natalka [10]

Hydrogen is a nonmetal and is placed above group in the periodic table because it has ns1 electron configuration like the alkali metals. However, it varies greatly from the alkali metals as it forms cations (H+) more reluctantly than the other alkali metals.Aug 15, 2020
Electronic Configuration: 1s
At Room Temperature: Colorless & Odorless Diatomic Gas
Atomic Weight: 1.0079

6 0

2 years ago

A GFCI is an example of what kind of electrical safety device?

vichka [17]

Answer:

Circuit breaker.

Explanation:

3 0

3 years ago

Read 2 more answers

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

<u>Where:</u>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

4 0

3 years ago