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2 years ago

Plz help I need to bring up my grade, this will help bring up my grade !!!!!!!

Chemistry

1 answer:

BartSMP [9]2 years ago

3 0

Answer:

WHY: You can abbreviate an element's electron configuration using the noble gas notation method because when you get down to the lower elements, specifically the d's and the f's, the electron configuration will be very long. The noble gas notation method is a faster answer while also being correct.

HOW: We can abbreviate an element's electron configuration by finding the last noble gas a specific element passed, for example calcium would have just passed Argon. Once you have the "address" of the previous noble gas, then you add on the difference between the element chosen and the noble gas, for example calcium would be [Ar] 4s^2.

Explanation:

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Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block.

SSSSS [86.1K]

The kinetic energy formula is;

$KE=\frac{m.v^2}{2}$

The variable $m$ represents the mass. Its unit is kilogram. We are informed that the mass of the object is $15kg$ . The variable $v$ represents the linear velocity. Its unit is meter per second. We are informed that the linear velocity of the object is $20m/s$ . Let's find the kinetic energy of the object by substituting these values in the formula.

$KE=\frac{15kg.(20m/s)^2}{2}$
$2.KE=15kg.400m^2/s^2$
$2.KE=(6000)kgm^2/s^2$
$KE=(3000)kgm^2/s^2=3000J$

5 0

11 months ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}