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6. What is the mass, in grams, of 1.92 moles of Ca(OH),?

Answer:

I don't get it is it even a question?

Answer:

Explanation:

From the information given;

Consider using Lande's Interval rule which can be expressed as:

[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]

here;

[tex]j+1[/tex]  = highest level of j

and

[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]

[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]

[tex]5(j+1) = 3(j+2)[/tex]

[tex]5j+5 = 3j+6[/tex]

[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]

recall that:

[tex]j = |S-L| \ \to \ |S+L |[/tex]

So;

[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &

[tex]S+L = \dfrac{5}{2} --- (1)[/tex]

Using the elimination method, we have:

[tex]2S = \dfrac{6}{2}[/tex]

[tex]S = \dfrac{3}{2}[/tex]

Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)

[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]

[tex]L = \dfrac{2}{2}[/tex]

[tex]L = 1[/tex]

Answer:

[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].

Explanation:

Identify the elements with oxidation state changes:

Oxidation states of iron, [tex]\rm Fe[/tex]:

Oxidation state of manganese, [tex]\rm Mn[/tex]:

The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:

[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].

(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)

Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].

Find the unknown coefficients using the conservation of atoms.

Reactants:

Therefore, among the products:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].

Products:

Reactants:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].

Products:

Therefore, among the products:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

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Answer:

movement of charged particles through the wire .

Explanation:

When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

   So;

    0.57mole of hydrogen gas will also produce 0.57 water

Answer:

In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M

Explanation:

In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10

In the first rinse the concentration must be of 0.9M  10 = 0.09M

2nd = 0.009M

3rd = 0.0009M

4th = 0.00009M

5th = 0.000009M →

Answer:

A. This allows the plants to conserve water and not bloom during the heat of the day

Explanation:

Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.

Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.

Answer:

Answer:

Reversible

Explanation:

Changes of state are physical changes in matter. Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization.

Answer: 7.77 g/ml

Explanation:

Volume of cylinder with only water = 3.28 mL

Volume of cylinder with water and metal = 8.72 mL

Volume of metal = (Volume of cylinder with water and metal ) -(Volume of cylinder with only water)

=8.72-3.28

=5.44 ml

Mass of metal = 42.26 g

Formula of Density =  [tex]\dfrac{\text{Mass}}{\text{Volume}}[/tex]

i.e. the density of the metal = [tex]\dfrac{42.26}{5.44}\approx7.77\text{ g/ml}[/tex]

Hence, the density of metal = 7.77 g/ml

Answer:

Explanation:

Nitric oxide is the gas NO, it reacts with oxygen as shown below;

2NO(g) + O2(g) -----> 2NO2(g)

Now the gas formed is the gas NO2 which is known to be reddish brown in colour.

A diagrammatic representation of this reaction is shown in the image attached to this answer.

Image credit: Chemlibretext

Answer:

Ideal Gas

The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.

Real Gas

The molecules of real gas occupy space though they are small particles and also have volume.

anation:

The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.

The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.

An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.

On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.

In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.

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Answer:

hello your question is incomplete attached below is the complete question

answer :

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

Explanation:

Reason for the mechanism

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

attached below is the detailed mechanism

Answer:

[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]

Explanation:

Hello!

In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:

[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]

Best regards!

Answer:

B

Explanation:

Answer:

0.696 atoms of oxygen

Explanation:

We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:

Mass of CaWO₄ = 50 g

Molar mass of CaWO₄ = 40 + 184 + (4×16)

= 40 + 184 + 64

= 288 g/mol

Mole of CaWO₄ =?

Mole = mass / Molar mass

Mole of CaWO₄ = 50 / 288

Mole of CaWO₄ = 0.174 mole

Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:

1 mole of CaWO₄ contains 4 atoms of oxygen.

Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.

Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.

Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.

Answer: The enthalpy change for this reaction is, -362.8 kJ

Explanation:

The balanced chemical reaction is,

[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]

[tex]\Delta H=-362.8kJ[/tex]

Therefore, the enthalpy change for this reaction is, -362.8 kJ

Answer:

c

Explanation:

ok than not c than b maybe

Answer:

25.8 g/dL

Explanation:

A chemist prepares a solution of aluminum sulfate by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Convert "V" to dL

We will use the following conversion factors.

450. mL × 1 L/1000 mL × 10 dL/1 L = 4.50 dL

Step 3: Calculate the concentration (C) of aluminum sulfate if g/dL

We will use the following expression.

C = m/V = 116.0 g/4.50 dL = 25.8 g/dL

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

The ocean is not a part of Earth's layers.

Answer:

Ocean

Explanation:

Answer:

2.9moles of hydrogen gas

Explanation:

convert liters to dm³

since 1liter= 1dm³

thus, 65.0liters = 65.0dm³

number of moles = volume given/22.4dm³

= 65.0/22.4

=2.9moles

Answer:

No

Explanation:

This does not mean that vinegar is insoluble totally. In fact, vinegar is soluble in water because water is a polar solvent.

For a substance to be soluble in another, it must obey the rule of solubility.

The rule states that "like dissolves like"

It implies that polar solvent will only dissolve polar solute.

Also, non-polar solvent will only dissolve non-polar solute.

Therefore, the answer is no, vinegar will dissolve in water.

Answer:

Lead

Explanation:

The subatomic particles within an atom can be used to know the atom or element given.

Of particular interest is the number of protons within the atom.

The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.

So; If we know the number of protons within an atom, we can know the element.

The number of protons given is 82, the element is  therefore lead.

Answer:

The atomic number of polonium is 84. The atomic number lead is 82.

Explanation:

Answer:

See explanation

Explanation:

The valence electrons are electrons found on the valence (outermost) shell of an atom.

When an atoms form compounds, there is an exchange of valence electrons between the atoms of one element and the atoms of another element.

Let us consider a typical example, sodium has one valence electron and chlorine has seven valence electrons. This means that chlorine needs one electron to complete its octet while sodium needs to release one electron in order to attain the octet structure.

So, sodium gives out its one electron and becomes a stable sodium ion and chlorine accepts that electron and becomes a stable chloride ion. This is how the compound sodium chloride is formed.

Answer:

60.0mL of the diluted solution are needed

6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.

Explanation:

As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:

3 * 20.0mL = 60.0mL of the diluted solution are needed

Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:

1.00M / 0.100M = 10 times must be diluted the solution.

As we need at least 60.0mL, the minimum volume of the stock solution must be:

60.0mL / 10 times =

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻     The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻   The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

Hope this helps

Strong acids and bases are those which completely ionized in body fluid, and weak acids and bases are those who does not completely ionized in body fluid.

Ka expression is used to differentiate between strong and weak acids.

[tex]\bold{\dfrac{[H+][F-]}{[HF]}}[/tex]

[tex]\bold{\dfrac{[NH_4^+] [OH^-]}{[NH_3]} }[/tex]

Given, [tex]\bold{ [H^+]=1\times10^-^6\; at \;25^oC}[/tex]

We know, [tex]\bold{ [H^+]\times[OH^-]=1\times10^-^6\; at \;25^oC}[/tex]

[tex]\bold{[OH^-]=\dfrac{1\times10^-^1^4}{6.2\times10^-^6} = 1.43\times10^-^9}[/tex]

Now, multiplying the value by [tex]10^1^0[/tex]

[tex]\bold{( 1.429\times10^-^9) \times 1\times10^1^0= 14.29}[/tex]

Thus, the value is 14.29.

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