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3 years ago

What methods do scientists use to determine if a change was physical or chemical?

1 answer:

5 0

A physical change will not change the chemical make-up of the material. the change will not produce a new substance.

a chemical change will have a reaction that produces heat, light, a change in color, releases a gas, or rust

You might be interested in

What is the minimum number of moles of potassium permanganate that would be required to completely oxidize 1.0 moles of benzalde

makvit [3.9K]

Mole ratio :

5 C₆H₆CHO + 2 KMnO₄ + 6 H⁺ = 5 C₆H₆COOH + 2 Mn²⁺ + 3 H₂O + 2 K⁺

5 moles C₆H₆CHO ------------------ 2 moles KMnO₄
1.0 moles C₆H₆CHO ---------------- ( moles of KMnO₄ )

moles of KMnO₄ = 1.0 x 2 / 5

moles of KMnO₄ = 2 / 5

= 0.40 moles of KMnO4

hope this helps!

7 0

3 years ago

Is the mass of .00566 mol of germanium (Ge) greater than one gram or less than one gram?

Salsk061 [2.6K]

Answer:

Less than one gram

Explanation:

Since there is no whole number before the decimal it means that the number is less than whole meaning it is less than one gram

8 0

2 years ago

Which of the following gases will have the highest velocity at a given temperature? (3 points)

grandymaker [24]

The correct option is C.
Helium gas will have the highest velocity at a given temperature because it has the lowest atomic mass of all the gases given. The atomic mass of gases affect their rate of movement. The lower the atomic mass, the faster the gas molecules moves and the higher the atomic mass, the slower the gas molecules move.

3 0

3 years ago

Read 2 more answers

A population of rabbits triples every 2 months If there are 2 rabbits initially, how long will it take for the population to inc

Bezzdna [24]

Answer:

The rabbit population will reach 500 after 10 months.

Explanation:

According to the given data:

The initial number of rabbit's equals 2.

Number of rabbit's after 2 months =2x3= 6

Number of rabbit's after 4 months = 6x3=18

Number of rabbit's after 6 months = 18x3=54

Number of rabbit's after 8 months = 54x3=162

Thus we can see that the number of rabbit's form a Geometric series with common ratio =3 and initial term = 2

Now the general term of a geometric series with first term 'a' and common ratio 'r' is given by

$T_{n+1}=ar^{n}$

Thus we need to find when the term becomes 500 thus using the given data we have

$500=2\cdot 3^{n}\\\\3^{n}=250\\\\(n)log_3(3)=log_3(250)\\\\(n)=5.025\\\\$

Thus the fifth term (excluding the start term) will have a rabbit count of 500 now since each term has a time difference of 2 months thus sixth term will occur after $5\times 2=10months$

7 0

3 years ago

In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat

Fiesta28 [93]

Answer:

Part A: (1, 1, 4, 1, 1, 1)

Part B: (2, 6, 4, 2, 3, 8)

Explanation:

Redox reactions can be balanced using the half-reaction method. It has the following steps:

We write both half-reactions (reduction and oxidation)
We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.
We add electrons to balance electrically the half-reaction
We multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.
We add both half-reactions and take the numbers to the general equation.

Acidic solution

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Basic solution

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻

4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.

6 0

3 years ago