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3 years ago

WILL GIVE BRAINLIEST ANSWER!!!

Chemistry

1 answer:

prohojiy [21]3 years ago

5 0

Answer: 2.07 M

Explanation:

(1.5m) (345ml)

---------------------

250 ML = 2.07M

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A covalent bond between an oxygen atom and a hydrogen atom is best described as _____.

Igoryamba

The answer is most likely nonmetals. :)

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3 years ago

Please help me vote you brainiest

bekas [8.4K]

Explanation:

An object in motion stays at motion An object at rest stays at rest unless acted by an external force.

[ Newton's 1st law of motion ]

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3 years ago

Is carbon dioxide a reactant

Harman [31]

Answer: No

Explanation: Reactants are the substances present at the beginning of a chemical reaction. In the burning of natural gas, for example, methane (CH4) and oxygen (O2) are the reactants in the chemical reaction. Products are the substances formed by a chemical reaction. In the burning of natural gas, carbon dioxide (CO2) and water (H2O) are the products formed by the reaction.

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3 years ago

How many moles are in 20 grams of argon

Sladkaya [172]

Answer:

There are 0.5 mole in 20g of argon.

Explanation:

40 g of argon = 1mole

Then 20g of argon is,

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5 0

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,

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

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

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

$E_a=22.87\ kJ/mol$

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

6 0

3 years ago