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2 years ago

Can a element be decomposed by chemical means

1 answer:

7 0

No, an element is a pure substance, it cannot be broken down into simpler substances or be decomposed by chemical means.

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If a position time graph has a constant slope, what can you infer about the<br> acceleration?

gogolik [260]

Answer:

Acceleration is zero.

Explanation:

The slope of a position time graph gives the velocity of the body.

If the slope is constant means the velocity is constant.

Now, acceleration is the measure of the change in velocity of a body over a given time interval.

So, the acceleration of a body is directly proportional to the change in velocity of the body.

If there is no change in velocity, this means that the acceleration of the body is zero.

Here, the slope is a constant implying that the velocity is a constant. So, there is no change in velocity. This implies that the acceleration is zero for the body in the given time interval.

Thus, if a position time graph has a constant slope, one can infer that the acceleration is zero.

5 0

2 years ago

Hg-178 undergoes 2 alpha decay and positron emission. Determine the new isotope.

Illusion [34]

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4 0

3 years ago

Hydrogen chloride, HCl, is classified as an Arrhenius acid because it produces(1) H+ ions in aqueous solution(2) Cl– ions in aqu

frozen [14]

Answer is (1) Produces H+ in aqueous solution

8 0

3 years ago

A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man

kap26 [50]

<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

Mass of NaBr sample is 11.97 g

% composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

A sample of 11.97 g of NaBr contains 22.34% of Na by mass

Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

= 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0

2 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago