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sweet-ann [11.9K]
2 years ago
10

Hello :) how to do this?

Chemistry
2 answers:
Nina [5.8K]2 years ago
7 0

Answer:

i dk

Explanation:

juin [17]2 years ago
4 0

Answer:

28g/ dm³

0.5 M

Explanation:

1dm³= 1000cm³

Let's convert the volume to cm³.

Volume of solution

= 200cm³

= (200÷1000) dm³

= 0.2 dm³

Concentration in g/dm³ means the amount of solute, KOH, in 1 dm³ of solution.

In 0.2 dm³, there is 5.6g of KOH.

Thus in 1 dm³, amount of KOH

= 5.6 ÷0.2

= 28g

Concentration in g/dm³ is 28g/dm³.

Molarity is a concentration unit. It is the number of moles of solvent in 1 L or 1dm³ of solution. Since we have already found the mass of KOH in 1dm³ of solution, we will use dm³.

1dm³ ----- 28g of KOH

moles= mass ÷mr

Molecular mass of KOH

= 39 +16 +1

= 56

1dm³ ----- 28 ÷56= 0.5 moles of KOH

Thus, concentration in molarity is 0.5M.

(Unit for molarity is M)

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erik [133]

Answer: Aluminium

Explanation: Aluminium metal has a lower density than copper. So, for the same volume of metal used to build a model airplane, the aluminium plane would be very lightweight while that of copper would be heavy.  The lightweight airplane will fly easily.

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2 years ago
Covalent compounds are generally not very hard. Justify the statement.
Eduardwww [97]

Covalent compounds are generally not very hard because they are formed by two or more nonmetallic atoms.

<h3>COVALENT COMPOUNDS:</h3>

Covalent compounds are compounds whose constituent elements are joined together by covalent bonds.

Covalent bonding occurs when two or more nonmetallic atoms of an element share valence electrons. This means that covalent compounds will not be physically hard since they constitute non-metals.

Examples of covalent compounds are:

  1. H2 - hydrogen
  2. H2O - water
  3. HCl - hydrogen chloride
  4. CH4 - methane

Learn more about covalent compounds at: brainly.com/question/21505413

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2 years ago
1. How many moles of LiBr are present in 100 mL of 1.25M LIBr solution?
Svet_ta [14]

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https://www.webassign.net/question_assets/wertzcams3/appendix.pdf

7 0
2 years ago
If you put an egg on a sidewalk on a hot day in July and try to cook it you are using?
Zepler [3.9K]

Answer:

  • <em><u>Passive solar energy</u></em>

Explanation:

First of all, you must know that you if you put an egg on a sidewalk you are dealing with energy from the Sun, i.e. solar energy, while geothermal energy is energy that comes from the inner of the Earth and biomass energy comes from plant or animal material.

The term passive solar energy refers to the fact that the energy of the sun is used directly for the intended task, which in this case is to cook the egg.

The term active solar energy refers to the fact that the energy of the Sun is converted into a  different form of energy and then used for your purpose. For instance, if the energy of the Sun were used to produce electricity and then this electricity used to cook the egg, you would be using an acitve solar energy.

5 0
3 years ago
A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0
2 years ago
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