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drek231 [11]
3 years ago
7

In one sample of a compound of copper and oxygen, 3.12g of the compound contains 2.50g of copper and the remainder is oxygen. Ca

lculate the percent composition for each compound
Chemistry
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

% composition O = 19.9%

% composition Cu = 80.1%

Explanation:

Given data:

Total mass of compound = 3.12 g

Mass of copper = 2.50 g

Mass of oxygen = 3.12 - 2.50 = 0.62 g

% composition = ?

Solution:

Formula:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition Cu = (2.50 g / 3.12 g)×100

% composition Cu = 0.80 ×100

% composition Cu = 80.1%

For oxygen:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition O = (0.62 g / 3.12 g)×100

% composition O = 0.199 ×100

% composition O = 19.9%

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Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
A 0.300 m solution of hcl is prepared by adding some 1.50 m hcl to a 500 ml volumetric flask and diluting to the mark with deion
liberstina [14]
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
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M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
8 0
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Carbohydrates are formed by plants converting water a d carbon dioxide into glucose and oxygen, in the photocatalyzed process is
Dmitry [639]
Photosynthesis maybe.
3 0
3 years ago
Polonium is a rare element with 33 radioisotopes. The most common one, 210Po, has 84 protons and 126 neutrons. When 210Po decays
Fofino [41]

Answer:

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Explanation:

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Now, we polonium has a proton number of 84 and a mass number of 210. Subtracting 2 and 4 respectively from its proton and mass numbers will yield 82 and 206 proton and mass numbers respectively.

Hence, the decomposition of the Po-210 isotope will yield an element with 82 proton number and 206 mass number. This corresponds to the element Lead.

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6 0
3 years ago
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AVprozaik [17]

Answer:

Check Explanation.

Explanation:

When we are talking about the reservoirs of carbon we mean the places where carbon are been stored. So, we will be considering the places where carbon can be stored in our planet. Remember that carbon is the 6th element in the periodic table or chart.

Within these Reservoirs, the carbon moves from one to another and that is called Carbon cycle. The reservoirs of carbon can be in the atmosphere, the lithosphere and in the biosphere.

(1). Carbon can be stored in the atmosphere as Carbondioxide,CO2.

(2). Carbon can be found on lithosphere in sedimentary rock such as limestone. Also, as components in fossil fuels.

(3). Carbon can be found in the biosphere as soil carbon.

5 0
3 years ago
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