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2 years ago

an alloy contains 56g of pure silver and 22g of pure copper what is the percentage(%) of silver in the alloy?

Chemistry

2 answers:

Lerok [7]2 years ago

6 0

To get the ans divide 5600by 78 and there is ur answer

gizmo_the_mogwai [7]2 years ago

4 0

<u>Answer:</u> The percentage of silver in the alloy is 71.8 %.

<u>Explanation:</u>

To calculate the percentage composition of a substance in a mixture, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

We are given:

Mass of pure silver = 56 g

Mass of pure copper = 22 g

Mass of alloy = 56 + 22 = 78 g

Putting values in above equation:

$\%\text{ composition of silver}=\frac{56}{78}\times 100=71.8\%$

Hence, the percentage of silver in the alloy is 71.8 %.

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Kaia, a chemical engineering graduate, has documented all titration procedures in her project report. She refers to this report

vagabundo [1.1K]

Answer:

The correct option is;

d. Explicit knowledge

Explanation:

Explicit knowledge is the knowledge that can be easily articulated documented stored in a retrieval system accesses, transmitted and shared with others

Tacit knowledge is the skill developed by an individual based on actual experience such that such knowledge comprise of both facts and perspectives

Hence explicit knowledge and tacit knowledge are complementary

The operations performed by Kaia include documentation, storing in a retrieval system (her project report) and accessing what she documented, this is an example of explicit knowledge.

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2 years ago

An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

jekas [21]

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

For example, $O_{2}$ , $N_{2}$ etc are pure substances.

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

pure chemical substance.

heterogenous mixture.

mixture.

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2 years ago

An ion has 24 protons and 19 electrons. Write the symbol for this ion.

Lerok [7]

Answer:

Explanation:

it hs 5 more protons thant electrons, so it has a positive charge of 5

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2 years ago

On a trip to the natural history museum you find two minerals that are similar in color. You can see from their chemical formula

lianna [129]

Answer:Yes they are in the same mineral group

Explanation:zinc is the central elements there. The rest of the elements are present as impurities due to where it was found. Like carbon is can be found in the soil, silicon with oxygen is basically sand, hydrogen is in the atmosphere and also in water and soil too. So apart from zinc, the rest are normal day to day elements.

7 0

2 years ago

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2

sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

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2 years ago