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2 years ago

Bronze contains 90 to 95 percent copper and 5 to 10 percent tin. Because these percentages can vary, bronze is classified as

2 answers:

8 0

Answer: a mixture.

Justification:

1) Pure substances have a definite chemical formula: the same kind of atoms with the same fixed ratios and chemical bonds. Therefore, the percents of each element do not varye.


2) Elements and compounds are pure substaces. For example, Fe, Mg, Ti, are elements, and CO₂, CO, H₂CO₃ are compounds. Each of them will have always the same kind of atoms, in the same ratio and with the same chemcial bonds. Therefore the percents of the elements do not varye.


3) Mixtures are formed by the physical combination (not chemical bonds) of different elements or compounds in variable proportions. As indicated, this describes the material bronze, in virtue of the variation of its composition. Other examples of mixtures are solutions (like brines), air, ocean water, and milk: different brines, different oceans and different milk have different contents of elements or compounds.

san4es73 [151]2 years ago

4 0

Answer:

it is a mixture

Explanation:

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A high school class is discussing whether a particular discovery in chemistry has had a positive impact on society. Which questi

AURORKA [14]

The most important question for the students to answer is what the discovery did to society. Did it change society in any way or better something?

3 0

2 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

$m=\frac{92.5}{100}\times 0.5940=0.54945 g$

1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

6 0

2 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

2 years ago

If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl

luda_lava [24]

I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{126580 g of benzyl chloride}{X}$

Solving for X,
X = $\frac{220 mg . 126580 mg}{353420 mg}$
X = 78.79 mg

For PPh₃:
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{262290 g of PPh3}{X}$

Solving for X,
X = $\frac{220 mg . 262290 mg}{353420 mg}$
X = 163.27 mg

Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when $\frac{95 percent}{100 percent}$ = $\frac{220 g}{X}$
Solving for X,

X = $\frac{220 . 100}{95}$ = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{78.79 g of benzyl chloride}{X}$
Solving for ,

X = $\frac{231.57 . 78.79}{220}$ = 82.93 mg of Benzyl chloride

when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{163.27 g of PPh3}{X}$
Solving for ,

X = $\frac{231.57 . 163.27}{220}$ = 171.85 mg of PPh3

So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.

7 0

2 years ago

In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?

Vinvika [58]

Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.

Number of gold, $Au$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms (given)

From the formula of compound that is $KAu(CN)_2$ it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.

So, the number of atoms of each element is:

Number of potassium, $K$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms

Number of carbon, $C$ atoms in $KAu(CN)_2$ = $2\times6.66\times 10^{20}$ atoms = $13.32\times 10^{20}$

Number of nitrogen, $N$ atoms in $KAu(CN)_2$ = $2\times6.66\times 10^{20}$ atoms = $13.32\times 10^{20}$

Total number of atoms in $KAu(CN)_2$ = Number of gold, $Au$ atoms+Number of potassium, $K$ atoms +Number of carbon, $C$ atoms + Number of nitrogen, $N$ atoms

Total number of atoms in $KAu(CN)_2$ = $6.66\times 10^{20}+6.66\times 10^{20}+13.32\times 10^{20}+13.32\times 10^{20}$

Total number of atoms in $KAu(CN)_2$ = $39.96\times 10^{20}$ atoms

Hence, the total number of atoms in $KAu(CN)_2$ is $3.996\times 10^{21}$ atoms.

7 0

2 years ago