1. Illustrate and describe the sequence in which ten electrons occupy the five
orbitals related to an atom's d sublevel.

The strong bases are hydroxides of group 1 and 2 metals (ex. Sodium, Potassium) or larger group transition metals (Copper, Iron), or oxides of the same metals.

It follows from structure of strong bases that they are composed of highly electropositive metals of groups 1 and 2 and the hydroxide ion.

Additionally, transition metals may also form bases with the hydroxide ion or oxides of this transition metals.

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Answer:

Explanation:

When comparing the drops of oil and water, one thing I noticed was the shape. The water drop was more defined, whereas the drop of oil began to spread and was much flatter. This may be due to the waxy material, and how both oil and water react to the wax.

Answer:

b. nuclear fuel has a higher energy density than fossil fuels.

Explanation:

Nuclear fuel is a quantity of energy measured in joules; therefore, it has the same units as electrical and mechanical energy.

Answer:

All of the signals pass through the thalamus

Explanation:

Answer:

S 2.6 [2.58] & C is 2.6 [2.55]

Explanation:

Answer:

Activation energy :)

Explanation:

The activation energy starts the reaction! Without it, we would not find out the solution!

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A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.

2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)

The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.

In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.

0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃

There are 1.25 g of KHCO₃ in the 1.775 g sample.

%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

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The medium that might be used to cool the hot liquid coming from a reactor is : ( C ) Heat exchanger

Heat exchanger is used  for the transfer of heat between liquids to be used in a chemical reaction, It can be used for both Cooling and heating to achieve the desired temperature.

The liquid coming from a reactor which posses a very high temperature will be passed through the heat exchanger so that its temperature will attain the required temperature for the experiment.

Hence we can conclude that The medium that might be used to cool the hot liquid coming from a reactor is Heat exchanger.

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Answer:

Water (H2O) and Sodium (Na) are reactants.

Explanation:

2H2O+2Na......2NaOH+H2

In the above reaction the water (H2O) and sodium (Na) are reactants and sodium hydroxide (NaOH) and hydrogen are products.

And this is the single displacement reaction.

Answer: D. 7.71 x 10^24

Explanation:

12.8 moles CO2 (6.02 * 10^23 molecules / 1 mol) = 7.7 * 10^24 molecules :)

The amount of heat energy lost by the 100 mL of water when it all freezes is 33400 J

Mass = Density ×Volume

Mass of water = 1 × 100

Mass of water = 100 g

Q = m × ΔHf

Q = 100 × 334

Q = 33400 J

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Answer:

Reflection

Hope it will help you

Answer:

(2) OH- is the answer.

Explanation:

Hope this helps:)...if not then sorry for wasting your time and may God bless you:)

The ions that combine with Ba2 to form the compound most soluble in water is OH- ion.

Barium is a chemical element with atomic number 58.

Barium reacts with the hydroxide ion to form barium hydroxide. It is also called baryta water.

The reaction is [tex]Ba^2^+ aq + 2OH^- \rightarrow Ba(OH)_2 (aq)[/tex]

Thus, the ions that combine with Ba2 to form the compound most soluble in water is OH- ion.

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Answer:

Polyester

Explanation:

We can store the copper sulphate solution in alumiun container, if cover on alumiun is present.

Aluminium is more reactive than copper so the Aluminium will displace copper sulphate from its solution by reacting with it but if there is cover on the aluminium then the alumium can't react with copper.

So we can store the copper sulphate solution in alumiun container.

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No, we cannot store CuSO₄ in an aluminum container because aluminum will displace the copper present in the solution.

Different chemicals are put in different containers, usually glass, plastic, dark container, or light container.

Chemicals element are reactive to different substances and to each other.

That's why they are put in separate and different types of containers according to their reactivity.

Thus, No, we cannot store CuSO₄ in an aluminum container because aluminum will displace the copper present in the solution.

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This is because amino group of p-aminobenzoic acid is an aniline and is less electrophilic than an alkyl amine.

This is an aromatic amine which consists of a phenyl group attached to an amino group.

The amino group of p-aminobenzoic acid being an aniline makes it less electrophilic which is why an alkyl amine participates in the reaction instead.

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The amino group of p-aminobenzoic acid does not participate in the reaction as this is an aniline and is less electrophilic than an alkyl amine.

Aniline is an aromatic amine molecule in which phenyl group is attached with an amino group.

The amino group present in the p-aminobenzoic acid behave as an aniline as a result of which it becomes less electrophilic in nature and this is the reason why an alkyl amine participates in the reaction instead of the p-aminobenzoic acid.

Hence amino group of p-aminobenzoic acid is less electrophilic in nature and that's why not participate in the reaction.

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Answer:

covalent bond- a chemical bond that is the sharing of electron pairs between atoms

The mass of nickel, Ni deposited when 0.2 faraday of electrical charge is passed through a solution of NiSO₄ is 5.9 g

Ni²⁺ + 2e —> Ni

Molar mass of Ni = 59 g/mol

Mass of Ni from the balanced equation = 1 × 59 = 59 g

Number of faraday = 2

SUMMARY

From the balanced equation above,

2 faraday deposited 59 g of Nickel

From the balanced equation above,

2 faraday deposited 59 g of Nickel

Therefore,

0.2 faraday will deposit = (0.2 × 59) / 2 = 5.9 g of Nickel

Thus, 5.9 g of Nickel were deposited in the process

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Answer:

5.9 grams

Explanation:

You need 2 faraday to deposit 1 mole of Ni (59 grams) because there are 2 moles of electrons in equation:

Ni+2 + 2e- --> Ni(s)

So, 0.2 faraday will deposit 1/10 of that or 5.9 grams of Ni(s).

Answer:

Club mosses

Explanation:

hope this helps

The formal charge on each atom can be known by counting up the number of valence electrons that is present on each atom.

The term Lewis structure is the representation of a compound using its symbols and showing the valence electrons as dots. This question is incomplete as the Lewis structures were not shown.

If it is shown, we can know the formal charge on each atom by counting up the number of valence electrons that is present on each atom.

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Answer:

The visual seems to be showcasing an erupting volcano, so I don't believe the two are related.

Explanation:

Answer:

Explanation:

The relation between molecular (formula) mass and molar mass page 44 . To obtain one mole of copper atoms (6.20 x 1023 ne atoms) weigh out 63.55 g copper. The molar mass (M) of a substance is the mass of one mole of entities (atoms, molecules, or formula units) of the substance.

Explanation:

Answer:

image 3 and 4show chemical reactions

Explanation:

For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.

n(Cd) / n(Fe)=1

Generally, the equation for the density  is mathematically given as

d=\frac{A}{4/3}\piR^3

Therefore

n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3

n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3

n(Cd) / n(Fe)=1

In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same

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Compared to the nucleus ⁵⁶₂₆Fe, the density of the nucleus ¹¹²₄₈Cd is same.

Density of any nucleus will be calculated by using the below equation as:

ρ = m/V, where

m = mass of the atom

V = volume = 4/3πR³

For the given question calculation will be proceed as:

n(Cd) / n(Fe) = [m(Cd)/m(Fe)].[R(Fe)/R(Cd)]³

From the given formulas of iron and cadmuim we put values on the above equation then we get,

n(Cd) / n(Fe) = (112 / 56 ) × (1/1.26)3

n(Cd) / n(Fe) = 1

Hence the density of nucleus ¹¹²₄₈Cd is same as of nucleus ⁵⁶₂₆Fe nucleus.

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Answer: test two

Explanation: it sits in between both

Answer:

1.12x10-10 or 0.63 ewan

Explanation:

Correct option is C)

The distance of centre of mass from carbon atom is,

xcm=m1+m2m1x1+m2x2

=12+1612×0+16×1.1

=0.63A˚

Thus, the centre of mass is 0.63A˚ from carbon atom.

The distance between a carbon atom and an oxygen atom in the co molecule is 0.63

Distance is the measurement of the farness of two objects from each other.

The distance between the molecules can be calculated by the given formula

[tex]xcm = m_1+m_2m_1\times 1+m_2 \times 2[/tex]

Putting the values in the equation

[tex]xcm = 12+1612 \times 0+16 \times 1.1 = 0.63 A^\circ[/tex]

Thus, the distance between a carbon atom and an oxygen atom in the co molecule is 0.63

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Answer:

Titrant and analyte is a pair of acid and base. ... Equivalence point: point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water.

Explanation:

Answer:

Decomposition reaction

Explanation:

When a single reactant is decomposed into two or more products that reaction is called decomposition reaction.

Option B is correct