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3 years ago

Explain what happens to substances during a chemical change

2 answers:

8 0

In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change.

icang [17]3 years ago

6 0

Answer:

A chemical change is when substances react chemically to make a product with a different identity than the reactants. The different identity is due to the rearrangement of the atoms in the reaction.

Explanation:

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The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van

Goshia [24]

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: Δ $T_{b}$ =i $K_{b}$ .m, where Δ $T_{b}$ = elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

$K_{b}$ = Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i= $\frac{2.5}{0.512 X 1.83}$

i=2.66= 2.7 (approx.)

6 0

3 years ago

Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0

2 years ago

If you add enzyme to a solution containing only the product(s) of a reaction, would you expect any substrate to form?

Masteriza [31]

Answer:

Explanation:

Provided the reaction that leads to the formation of the products can proceed in both forward and backward directions, the correct answer would be yes because the reaction will proceed backward until equilibrium is reached.

For a reaction that can proceed both forward and backward, the addition of a catalyst increases the rate of reaction in both directions based on the fact that a catalyst cannot alter the equilibrium of a reaction.

Hence, if an enzyme is added to the product of a reaction that has the potential to proceed in both forward and reverse reactions, a substrate would be expected to form because the reaction will proceed backward until an equilibrium is reached.

The correct option is e.

8 0

2 years ago

How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

6 0

3 years ago

Why is Avogadro’s number referred to as a mole? If you could change this terminology, what would you change it to and why?

stepan [7]

The number of particles (molecules, atoms, compounds, etc.) per mole of a substances is known as Avagadro number. It is equal to 6.022×10^23 mol-1 and is expressed as NA.

Number of moles is the amount of a substance that contains as many particles as there are atoms in 12 grams of pure carbon-12. So, 1 mol contains 6.022×10^23 elementary entities of the substance. Since 6.022 x 10^23 is the Avagadro number, one mole is equal to Avagadro number.

One mole of a substance is the ratio of mass of the substance by the molecular mass of the substance. Thus the mass of one mole of a substance is equal to the substance's molecular weight. Thus one mole of a substance is the atomic mass unit of a substance and since one mole is equivalent to the Avagadro number,we can conclude that one Avagadro number is one atomic mass unit of the substance.

7 0

3 years ago