A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.

A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%

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

Answer:

Ocean

Explanation:

Answer:

(a) 1s; has one orbital

(b) 4d; has five orbitals

(c) 3p; has three orbitals

(d) n=3 has nine orbitals

Explanation:

Electrons in an atom are always in constant motion, making it hard to predict there exact position. However, the most probable locations electrons can be be found are described with the terms shells, subshells and orbitals. A shell contains subshells and orbitals are found within subshells. The shells are given names such as K, L, M, N, which correspond to the principal quantum numbers, n = 1, 2, 3, and 4 respectively. There are 4 major types of subshells that can be found in a shell. They are named as s, p, d, f. Each subshell is composed of several orbitals.

a. 1s; the s subshell has only one orbital. Therefore, the 1s subshell has one orbital

b. 4d; the d subshell has five orbitals. Therefore, the 4d subshell has five orbitals

c. 3p; the p subshell has three orbitals. Therefore, the 3d subshell has  three orbitals

d. n = 3; the shell with n = 3 has the following subshells, 3s, 3p, 3d.the number of orbitals will be 1 + 3 + 5 = 9 orbitals. Therefore, the number of orbitals in n = 3 is nine orbitals

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

Where [] are equilibrium concentrations of each specie

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

Answer:

d. 9.95 × 10⁻³ mol

e. 249 g

f. 2.55 × 10²⁴ atoms

g. 4.32 × 10²² atoms

Explanation:

d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol

e. the mass of 4.23 mol of cobalt

The molar mass of cobalt is 58.93 g/mol.

4.23 mol × 58.93 g/mol = 249 g

f. the number of cobalt atoms in 4.23 mol of cobalt

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms

g. the number of cobalt atoms in 4.23 g of cobalt

First, we will calculate the moles of cobalt using the molar mass of cobalt.

4.23 g × 1 mol/58.93 g = 0.0718 mol

Then, we will calculate the number of cobalt atoms using Avogadro's number.

0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms

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: product

Explanation:

Each substance written to the right of the arrow in a chemical equation is referred to as a product.

When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.

On the other hand which is the right side is the product.

Answer:

B

Explanation:

Answer:

c

Explanation:

ok than not c than b maybe

Answer:

H-H equation is written as follows:

pH=pK + log

{HCO3-}(base)

{H2CO3}(acid)

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:

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:

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:

th relative ability of a solute to devolve into a solvent

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:

Partial pressures:

PCl₅ = 0.558 atm

PCl₃ = 0.22 atm

Cl₂ = 0.22 atm

Explanation:

From the given information:

The number of moles of PCl₅ associated with the evaporation is:

[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]

[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]

[tex]n_{PCl_5}= 0.013 \ mol[/tex]

Temperature of the gas = 250° C = (250 + 273.15) K

= 523.15 K

Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅

PV = nRT

[tex]P = \dfrac{nRT}{V}[/tex]

[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]

P = 0.558 atm

Thus, at  250° C, decomposition of PCl₅ occurs.

In the container, PCl₅  decomposes to PCl₃ and Cl₂.

i.e.

[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]

Using Dalton's Law:

[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]

[tex]P_1 = P_{Total} \times X_1[/tex]

where;

X = mole fraction

Then, the total no. of moles in the container is:

[tex]n = \dfrac{PV} {RT}[/tex]

[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]

n = 0.023 mol

Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.

Thus, the partial pressure of  PCl₃  is:

[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]

[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]

[tex]P__{PCL_3}} = 0.22 \ atm[/tex]

Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm

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:

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: 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:

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:

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.

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:

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:

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:

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:

The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.

Explanation:

Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.

Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).

These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.

Therefore,  Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.

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Answer:

e. I, At

Explanation:

Hello!

In this case, since the periodic trends of a series of elements belonging to the same group towards physical and chemical properties tend to be the same when closer in period, we notice that Mg and Al, Br and Kr and As and Br are close but in period, not in the same group; therefore e. I, At, iodine and astatine, are going to tend to exhibit the greatest similarity in their physical  and chemical properties.

Best regards!

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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