1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tangare [24]
2 years ago
6

State which substance is undergoing oxidation and which substance is undergoing reduction in the following reaction:

Chemistry
1 answer:
yKpoI14uk [10]2 years ago
8 0

Answer:

Mg(s) = oxidized

H₂SO₄(aq) (hydrogen) = reduced

Explanation:

Oxidized substances lose electrons and see an increase in oxidation number. Reduced substances gain electrons and see a decrease in oxidation number.

Lone elements always have an oxidation number of 0. H₂SO₄ and MgSO₄ are ionic compounds that can be broken up into a cation and polyatomic anion. The anion, SO₄²⁻, has an overall oxidation number of -2. Therefore, to make the overall charge of the ionic compound neutral, the hydrogen in H₂SO₄ has an oxidation state of +1 (bc there are two hydrogens) and the magnesium in MgSO₄ has an oxidation state of +2.

Since Mg(s) is losing electrons (goes from 0 to +2), it is being oxidized. Since the hydrogen in H₂SO₄(aq) is gaining electrons (goes from +1 to 0), it is being reduced.

You might be interested in
1. Halogens are good disinfectants. Which of the following is a halogen?
katen-ka-za [31]

Answer:

C. Cl

Explanation:

6 0
3 years ago
How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper
satela [25.4K]

Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

Initial temperature = 25 °C

Final temperature = 75 °C

Energy absorbed = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 75 °C - 25 °C

ΔT = 50 °C

now we will put the values in formula

q = m . c . ΔT

q = 150 g × 4.186 J/g.°C × 50 °C

q = 31395 J

so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .

5 0
3 years ago
Calculate these volumes.
MrMuchimi
Check photo for answers!

7 0
3 years ago
Read 2 more answers
A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3 . calculate the density of gold.
Fantom [35]

Answer:

            19.29 g.cm⁻³

Solution:

Data Given:

                             Mass  =  301 g

                             Volume  =  15.6 cm³

Formula Used:

                             Density  =  Mass ÷ Volume

Putting values,

                             Density  =  301 g ÷ 15.6 cm³

                            Density  =  19.29 g.cm⁻³

6 0
3 years ago
In an experiment, the molar mass of the compound was determined to be 118.084 g/mol. What is the molecular formula of the compou
Mrrafil [7]

Answer:

need more info

Explanation:

i think so we need more info

4 0
2 years ago
Other questions:
  • What is the formula for Volume?
    5·1 answer
  • Which statement describes a chemical property of matter?
    6·2 answers
  • Be sure to answer all parts.The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, fro
    7·1 answer
  • For chemical reactions involving ideal gases, the equilibrium constant K can be expressed either in terms of the concentrations
    12·1 answer
  • Hello, please help me if you can :(
    8·1 answer
  • Describe what a mole is,<br>and what is meant by<br>a molar solution.​
    11·1 answer
  • A hurricane's surface winds are responsible for transporting water vapor into the
    13·2 answers
  • Transfer RNA __________. A is translated into protein B tags a growing polypeptide for export to the endoplasmic reticulum C is
    12·1 answer
  • What is the best explanation for why a particle is striking point X?
    12·2 answers
  • Which metal pan should the chef choose because it undergoes the largest temperature change for a given input of energy?
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!