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2 years ago

What happens to the total mass of a substance undergoing a physical change

2 answers:

7 0

That depends. What kind of change are you talking about? But

Mass through chemical change stays the same as well. Example: burning paper, the ash left behind is not all of the mass of the reactants, Carbon dioxide, and other substances also makeupmass but just is not seen

Lady_Fox [76]2 years ago

6 0

If your answer choices are:

It stays the same.

It doubles.

It triples.

It is reduced.

The correct answer is "It stays the same."

You might be interested in

The chemical reaction that causes chromium to corrode in air is given by 4Cr+3O2→2Cr2O3 in which at 298 K ΔH∘rxn = −2256 kJ ΔS∘r

MAVERICK [17]

Answer:

-2092 kJ

Explanation:

Let's consider the chemical reaction that causes chromium to corrode in air.

4 Cr + 3 O₂ → 2 Cr₂O₃

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ΔH° - T × ΔS°

where,

ΔH°: standard enthalpy of the reaction

T: absolute temperature

ΔS°: standard entropy of the reaction

ΔG° = -2256 kJ - 298 K × (-0.5491 kJ/K)

ΔG° = -2092 kJ

5 0

2 years ago

Read 2 more answers

If an ionic compound were composed of a4+ and b−, which unit cell structure would give a neutral compound?

Ymorist [56]

biological reactions that happen within cells while reducing the complex interactions found in a whole cell. Eukaryotic and prokaryotic cells have been used for creation of these simplified environments[1]. Subcellular fractions can be isolated by ultracentrifugation to provide molecular machinery that can be used in reactions in the absence of many of the other cellular components. Cell-free biosystems can be prepared by mixing a number of purified enzymes and coenzymes. Cell-free biosystems are proposed as a new low-cost biomanufacturing platform compared to microbial fermentation used for thousands of years. Cell-free biosystems have several advantages suitable in industrial applications

4 0

2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, ΔTf is the depression in freezing point of water.

i is van't Hoff factor.

Kf is the molal depression constant.

m is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).

van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0

2 years ago

After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ___

mamaluj [8]

Answer:

The answer to be filled in the respective blanks in question is

3 and 1

Explanation:

So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.

Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1

7 0

2 years ago