Copper(II) sulfate is an active ingredient in tree stump remover. In order to help remove

algol [13]

I’m not sure what kind of answer you’re looking for, but if you’re supposed to translate it into a sentence it would be two moles of sodium bicarbonate(baking soda) decompose unto one mole of sodium carbonate(soda ash) plus one mole of water and one mole of carbon monoxide gas

8 0

3 years ago

Carbon dioxide is a compound of the elements carbon and oxygen (CO2). What happened to the carbon and oxygen to create this comp

dexar [7]

Answer:

D. they exchanged electrons

6 0

2 years ago

HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

OleMash [197]

<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

<u>Explanation:</u>

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

$HNO_3+S\rightarrow H_2SO_4+NO$

<u>On the reactant side:</u>

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

<u>On the product side:</u>

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

3 0

3 years ago

Equal moles of H2, N2, O2, and He are placed into separate containers at the same temperature. Assuming each gas behaves ideally

lbvjy [14]

Answer:

They would all exhibit the same pressure.

Explanation:

Since the same number of mole of each gas is placed in different containers, it means the gas will occupy the same volume.

Now, the gases were observed at the same temperature. This means they will all have the same pressure as their volume is the same.

Now we can further understand this by doing a simple calculation as follow:

Assumptions:

For H2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, H2 has a pressure of 1 atm.

For N2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, N2 has a pressure of 1 atm

For O2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, O2 has a pressure of 1 atm

For He:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, He has a pressure of 1 atm.

From the above illustrations we can see that the gases have the same pressure since they have the same number of mole, volume and were observed at the same temperature.

4 0

3 years ago

5)

Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

$\tt \dfrac{18.24}{12}=1.52$

mol H(MW=1 g/mol) :

$\tt \dfrac{0.51}{1}=0.51$

mol F(MW=19 g/mol)

$\tt \dfrac{16.91}{19}=0.89$

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

6 0

3 years ago