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3 years ago

A sentence with the word atom

Chemistry

2 answers:

Sati [7]3 years ago

8 0

Answer:

we use atom in our sentences all the time

Explanation:

that what i put

faltersainse [42]3 years ago

8 0

We use atom in our sentences all the time and half the time we don’t even realize it!

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Answer:

Yes

Explanation:

because some foods have carbohydrates but are healthy

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3 years ago

5) How many moles are in 5 Liters of H2 gas?

telo118 [61]

From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
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3 years ago

How do you calculate the number of moles in CO2

gogolik [260]

Answer:

The number of molecules in a mole (known as Avogadro's constant) is defined such that the mass of one mole of a substance, expressed in grams, is equal to the mean molecular mass of the substance. The molecular mass of CO2 = 12+2x16 = 44, so the mass of a mole of CO2 is approximalty 44 grams

Explanation:

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3 years ago

How do the rates of the forward and reverse reactions change throughout a reaction?

fenix001 [56]

The rates of the forward and reverse reaction depends on the temperature on which the reaction will proceed, either endothermic of exothermic. it also depends of the concentration of the reactants and products. if the reaction is exothermic, so if the reaction temperature is increased then it will favor the forward reaction, then if the reaction is lowered then it will favor the reverse reaction

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3 years ago

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,

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

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

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

$E_a=22.87\ kJ/mol$

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

6 0

3 years ago