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2 years ago

In any chemical compound, the elements are always combined in the same proportion by ___. A. charge B. mass C. volume D. density

Chemistry

1 answer:

Tresset [83]2 years ago

6 0

Answer is: B. mass.

Proust's law or law of constant composition said that a given chemical compound always contains its component elements in fixed ratio and does not depend on method of preparation.

Law of multiple proportions or Dalton's Law said that the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.

For example, nitrogen(I) oxide N₂O; m(N) : m(O) = 2·14 : 16 = 7 : 4.

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After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

6 0

3 years ago

for the reaction 2Fe + O2 = 2FeO, how many grams of iron oxide are produced from 8.00 mol of iron? when o2 is an excess

Luba_88 [7]

Answer:

2Fe + O₂ -------------------> 2FeO

8 mol Fe produce

8 mol Fe * 2 mol FeO / 2 mol Fe = 8 mol FeO

Mass of FeO = 8 mol FeO * 71.85 g/mol = 574.8 grams FeO

Explanation:

Having 8 mol of Iron means 8 moles of iron oxide can be produced. Each mole of iron oxide has a molecular weight of 71.85 grams. Therefore, 8 moles of iron oxide should weight 574.8 grams.

7 0

3 years ago

Consider the reaction 5 Br− (aq) + BrO3− (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) The average rate of consumption of Br− is 1.8

kaheart [24]

Answer : The average rate of consumption of $H^+$ during the same time interval is, 2.17 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

As we are given:

$\frac{d[Br^-]}{dt}=1.81M/s$

Now we have to determine the average rate of consumption of $H^+$ during the same time interval.

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

or,

$-\frac{1}{6}\frac{d[H^+]}{dt}=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times (1.81M/s)$

$\frac{d[H^+]}{dt}=2.17M/s$

Thus, the average rate of consumption of $H^+$ during the same time interval is, 2.17 M/s

6 0

3 years ago

The Diagram Shows The Box For An Element In The Periodic Table What Is The Atomic Mass Of The Element Shown?

Talja [164]

Answer:

Explanation:

5 0

2 years ago

Read 2 more answers

How is a carbon atom different from a hydrogen atom?

Sloan [31]

Answer:

Carbon atom is a chemical element with the symbol C and atomic number 6 ( electrons). While Hydrogen atom is an atom of the chemical element hydrogen that have only 1 electron.

Explanation:

3 0

2 years ago