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3 years ago

We do we mean when we say a substance is pure

Chemistry

1 answer:

tiny-mole [99]3 years ago

6 0

Answer:

ot means that contains only one kind of matter.

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Identify the element: 6 energy level, metallic, 82 election

Lady bird [3.3K]

Answer:-

Name of element = Lead

Symbol ♾ of Lead = Pb

Atomic no. = 82

Atomic mass = 207.2 amu

No. of protons = 82

No. of electrons = 82

Yes, lead(Pb) is a metallic element and certainly it has 6 electron shells which means 6 energy level.

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3 years ago

How do you calculate atomic number

kifflom [539]

The number of protons in the nucleus of an atom is equal to the atomic number of an element. You can also find it by subtracting the number of neutrons from the atomic mass. Atomic Number = Atomic Mass - No. of Neutrons.

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3 years ago

Earth’s inner core:_____ as Earth’s outer core:liquid

KatRina [158]

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1 year ago

How does a catalyst increase the rate of a chemical reaction?

boyakko [2]

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3 years ago

Read 2 more answers

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

The activation energy for this reaction = 23 kJ/mol.

6 0

2 years ago