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2 years ago

A reduction reaction involves either the addition of hydrogen or removal of:

Chemistry

1 answer:

Marina CMI [18]2 years ago

4 0

Reduction involves the either the addition of hydrogen and removal of oxygen.

<h3>What is reduction?</h3>

Reduction involves the removal of oxygen.

This implies there is a loss of oxygen in reduction.

This can be represented in the extraction of iron from it ores.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Reduction is also the addition of hydrogen. This implies it is the gain of hydrogen.

For example

CH₃CHO → CH₃CH₂OH

learn more on reduction here: brainly.com/question/9485345

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Explain how our body systems work together to get oxygen into and around our body.

aleksandrvk [35]

Answer: (1) Inhales (breathes in) Oxygen - good for the body - gives it to the Circulatory System to be transported throughout the body through the blood. (1) Digestive System gets nutrients (good) from food and hands it over to the blood and Circulatory System then carries those nutrients where they need to go.

7 0

3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

3 years ago

A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 293 K at a pressure of

Mekhanik [1.2K]

Answer:

2.42L

Explanation:

Given parameters:

V₁ = 1.8L

T₁ = 293K

P₁ = 101.3kPa

P₂ = 67.6kPa

T₂ = 263K

Unknown:

V₂ = ?

Solution:

To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

All the units are in the appropriate form. We just substitute and solve for the unknown;

101.3 x 1.8 / 293 = 67.6 x V₂ / 263

V₂ = 2.42L

7 0

3 years ago

What is the mass in grams of 85.32 mL of blood plasma with a density of 1.03 g/mL

sveticcg [70]

<span>Well if you're looking for grams, all you need to do is cancel out units. (ml)(g/ml)=g because the ml cancels out. Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :) Good luck! Hope that helped. When in doubt, look at the units.</span>

3 0

3 years ago

Calculate the volume of hydrogen if you have 12.1 moles of Hydrogen.

vodomira [7]

To Find :

The volume of 12.1 moles hydrogen at STP.

Solution :

We know at STP, 1 mole of gas any gas occupy a volume of 22.4 L.

Let, volume of 12.1 moles of hydrogen is x.

So, x = 22.4 × 12.1 L

x = 271.04 L

Therefore, the volume of hydrogen gas at STP is 271.04 L.

8 0

2 years ago