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Trava [24]
2 years ago
15

Elements found on the left side of the Periodic Table are known as

Chemistry
1 answer:
Degger [83]2 years ago
5 0

Answer:

Metals

Explanation:

Akaline earth metals to be exact

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Choose the words to finish the sentence. After learning about the law of conservation of mass, Sammy became interested in balanc
cestrela7 [59]

The chemical equation is unbalanced and synthesized.

<h3></h3><h3>What is a chemical equation?</h3>

A chemical equation is described as the symbolic representation of a chemical reaction in the form of symbols and chemical formulas.

In a chemical equation, the reactant entities are given on the left-hand side and the product entities is shown on the right-hand side with a plus sign between the entities in both the reactants and the products, and an arrow that indicates towards the products to show the direction of the reaction.

We can conclude that in the chemical equation shown is unbalanced because both amounts of the individual elements and compounds do not reflect on the reactant and product side.

Learn more about chemical equations at: brainly.com/question/11231920

#SPJ1

The complete question is below:

After learning about the law of conservation of mass, Sammy became interested in balancing equations. He knew that the symbol for aluminum was Al and silver tarnish was Ag2S. He also knew that mixing the two chemicals yielded pure silver, or Ag, in an aluminum sulfide solution. Here is the equation showing this reaction:

3 Ag2S + 2 Al → 6 Ag + Al2S3

This equation is (synthesis / unbalanced / replacement / balanced), and it represents a(n) (unbalanced / balanced / synthesized / replaced) chemical reaction.

answer choices:

  • synthesis; balanced

  • balanced; replacement

  • unbalanced; synthesized

  • balanced; balanced

6 0
1 year ago
Classify each of the following particulate level illustrations as a representation of either a pure substance, a homogeneous mix
Anarel [89]

Answer:

The classification and illustrations are attached in the drawing.

Explanation:

It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.

On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.

4 0
3 years ago
Examples of a pure substances
DiKsa [7]

Answer:

Explanation:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.

3 0
2 years ago
Read 2 more answers
What is a non-example for an Electric circuit?
Mademuasel [1]

Answer:

Impressionist paintings are a non-example of electric circuits.

Explanation:

hope this helped

4 0
2 years ago
Read 2 more answers
A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
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