If an element is bonded to 4 other atoms and has a formal charge of +1, what group must the element be in?
A) 5A
B) 6A
C) 7A
D) 3A
E) 4A

Answer:

Kc = 0.101 or 1.01*10^-1

Explanation:

Set up ICE table
2NH3 -> 3H2 + N2
I: 2 moles, 0 moles, 0 moles

C: -x moles   +y moles   +z moles

E: 1.25 moles 0+y moles   0+z moles

We know that x is 0.75 moles, since 2-0.75 is 1.25

for every mole of NH3 consumed, 1.5 moles of H2 are produced, and 0.5 moles of N2 are produced. If 0.75 moles of NH3 are consumed, then 1.5 * 0.75 moles of H2 are produced, and 1.5*0.5 moles of N2 are produced

so, y equals 1.5 * 0.75, which is 3/2 * 3/4, or 3/8, or 0.375 moles of H2 produced

z equals 1.5 * 0.5, which is 0.75 moles of N2 produced

Kc equals [tex]\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]

N2 = 0.75, N2 = 0.375, and NH3 = 1.25
So, Kc equals
(0.375 * 0.75^3)/(1.25^2), which equals 0.10125

Round to 3 sig figs to get 0.101

Calcium sulfate, carbon dioxide, and water are produced when sulfuric acid and calcium carbonate interact.

The insoluble in water free lime (CaCO3) interacts with sulfuric acid. It takes only a few minutes for sulfuric acid and lime to react and produce gypsum. H2SO4 + Lime CaCO3 = CaSO4 (gypsum).

Each of these metals forms a layer of insoluble sulfate when exposed to sulfuric acid, which slows or stops the reaction altogether. A white precipitate of calcium sulfate and a small amount of hydrogen are formed in the calcium case.

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The concentration of the solution is obtained as 7.3 * 10^-6 M.

We know that one of the ways by which we are able to obtain the concentration of substances especially those that are colored is by the use of spectrophotometry. In a spectrophotometer, there is a wavelength of maximum absorption that could be used to measure the concentration of the solution as we can see in the question.

By the use of the Beer Lambert's law we have;

A = εcl

A = Absorption

ε = molar absorptivity

c = concentration

l = path length

Thus we would have;

0.2 = 27400 M-1cm-1 * c * 1.00 cm

c = 0.2/27400 M-1cm-1 *  1.00 cm

c = 7.3 * 10^-6 M

We would have from the calculation that we have a solution whose concentration is 7.3 * 10^-6 M.

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The theoretical yield  of the precipitate is  1185 g and the balanced reaction equation is; [tex]3Pb(NO_{3} )_{2} (aq) + Ni_{2} (SO_{4}) _{3} (aq) -------- > 2Ni(NO_{3} )_{3}(aq) + 3PbSO_{4}(s)[/tex]

We know that the term chemical reaction equation has to do with the interaction that exists between reactants and product. Now we know that a balanced reaction equation ought to have the same number of atoms of each of the elements that were combined in the reaction.

We have the balanced reaction equation as;

[tex]3Pb(NO_{3} )_{2} (aq) + Ni_{2} (SO_{4}) _{3} (aq) -------- > 2Ni(NO_{3} )_{3}(aq) + 3PbSO_{4}(s)[/tex]

Now;

Number of moles of nickel III sulfate = 531 g/406 g/mol = 1.31 moles

Number of moles of lead II nitrate = 3.91 moles

Since 3 moles of nitrate reacts with 1 mole of the sulfate

x moles of the nitrate reacts with 1.31 moles of the sulfate

x = 3.93 moles

Thus  lead II nitrate is the limiting reactant.

If 3 moles of the nitrate produces 3 moles of the precipitate then

Theoretical yield = 3.91 moles  * 303 g/mol

= 1185 g of precipitate

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The nomenclature name of NaNO is Sodium Nitrite or could also be said as Sodium Nitrate.

Sodium nitrite or sodium nitrate is a chemical compound with Sodium cation and nitrate anion. So the combination of these two ions gives us Sodium nitrite.

It is a white, ionic, solid crystalline compound that is used as a food preservative and additive.

Sodium nitrite and sodium nitrate are both inorganic compounds with the chemical formula NaNO₃.

They are both white, crystalline solids that are highly soluble in water. Sodium nitrite is the active ingredient in many curing agents used to preserve meats, and it is also used as a color fixative in smoked fish. Sodium nitrate is used as a fertilizer and as an oxidizing agent in pyrotechnics.

Therefore the nomenclature name of the chemical compound is Sodium Nitrite or Sodium nitrate depending upon the anion NO₂ or NO₃

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The heat absorbed by the water when A\a chunk of iron at 90.6°C was added to 200.0 g of water at 15.5°C in a coffee cup calorimeter is 2257.2 joules and the mass of iron added is 69.44 grams

We know, Heat gained = mass of sample * temperature change * specific heat

thus a)  heat gained by water = 200g *(18.2-15.5) °C * 4.18 J/g°C = 2257.2 Joules

also since no heat lost to the surrounding we can say,

b) Heat lost by the chunk of iron = heat gained by water = 2257.2 Joules

or, heat gained by iron = -2257.2 Joules

using the same formula for chunk of iron,

Heat gained = mass of sample * temperature change * specific heat

-2257.2 J = mass of iron * (18.2-90.6)°C * 0.449 J/g°C

mass of iron = 2257.2/( 72.4 * 0.449 ) = 69.44 grams

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Answer:

algebraic reaction is the answer of this question

0.0001212598 is a long will it be for one-half of carbon-14 in a sample to remain.

A reaction's half-life is the amount of time needed for the reactant concentration to drop to half its initial value. The rate constant for a first-order reaction and its half-life are connected by the constant t1/2 = 0.693/k.

The half-life of carbon-14 is 5,730 40 years, meaning that during the following 5,730 years, half of the radioisotope that is present at any particular time will spontaneously disintegrate.

Carbon-14's 5715-year half-life

The decay constant needs to be determined.

We know the half-life equation.

t[tex]\frac{1}{2}[/tex] = 0.693/λ

where,

t1/2  ​is the half-life and λ is the decay constant

so,

λ = 0.693 / 5715 = 0.0001212598

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The volume in liters of 5.75 moles of CO₂ at standard temperature and pressure (STP) is 128.88L.

The volume of a given gas at STP can be calculated using the following formula;

PV = nRT

Where;

According to this question, 5.75 moles of CO₂ at standard temperature and pressure (STP). The volume of this gas can be calculated as follows;

At STP, the following applies;

1 × V = 5.75 × 0.0821 × 273

V = 128.88L

Therefore, 128.88L is the volume of the gas.

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The percent yield for the reaction below when 364 g SO₂ and 42.0 g O₂ produce 408 g SO₃ is 51.5 %.

The reaction is given as :

2SO₂(g) + O₂(g)   ------>   2SO₃

mass of SO₂ = 364 g

mass of O₂ = 42 g

mass of SO₃ = 408 g

moles of SO₂= mass / molar mass

                     = 364 / 64

                     = 5.68 mol

moles of O₂ = mass / molar mass

                    = 42 / 32

                   = 1.31 mol

1 mole of O₂ react with = 2 mole of  SO₂

1.31 mole of O₂ react with = 2 × 1.31 = 2.62 mol of  SO₂

this means O₂ is limiting reagent.

1 mole of O₂  produce 2 mole of SO₃

1.31 mole of O₂  = 2 × 1.31 = 2.62 mol of SO₃

mass of SO₃ = 2.62 × 80

                     = 209.6 g

Percent yield =( actual yield / theoretical yield ) × 100 %

                      = (209.6 / 408) × 100 %

                      = 51.5 %

Thus, The percent yield for the reaction below when 364 g SO₂ and 42.0 g O₂ produce 408 g SO₃ is 51.5 %.

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Answer:

1.67262192 × 10-27 kilograms

The mass of a proton is 1.6726219 × 10-27 kilograms.

What do protons do?

Proton properties:

Proton Mass: One proton has the same mass as a hydrogen atom. A proton and a electron together make up a hydrogen atom. The mass of a hydrogen atom is equal to the mass of a proton because the mass of an electron is thought to be negligible. An electron weighs 1840 times as much as a proton.

Proton mass is equal to 1.676 x 1027 kilograms, or 1.676 x 1024 grams.

Charge of Proton - A proton's charge is equal to and in the opposite direction of an electron's. Because of this, it has a unit positive charge.

A proton has a charge of +1.602 10-19 coulombs.

Hence, The mass of a proton is 1.6726219 × 10-27 kilograms.

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Answer:

53.7 L

Explanation:

The volume occupied by 35.2 g of methane gas at 25 °C and 1.0 atm. is 55.9 L.

Given to us is temperature, pressure, universal gas constant, and mass of methane gas, we need to find the volume occupied by the gas.

To calculate the volume occupied by the methane gas, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.08206 Latm/(Kmol))

T = temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 25 °C + 273.15

T(K) = 298.15 K

Next, we calculate the number of moles of methane gas using its molar mass:

molar mass of CH₄ = 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol

n = mass/molar mass

n = 35.2 g / 16.04 g/mol

n = 2.19 mol

Now we can substitute the values into the ideal gas law equation:

PV = nRT

V = (nRT) / P

V = (2.19 mol × 0.08206 Latm/(Kmol) ×298.15 K) / 1.0 atm

V = 55.9 L

Therefore, the volume occupied by 35.2 g of methane gas at 25 °C and 1.0 atm is 55.9 liters.

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Taking into account the reaction stoichiometry, 11.2 liters of hydrogen gas can form at STP when 0.500 L of 2.00M HCl solution reacts with excess magnesium.

In first place, the balanced reaction is:

Mg + 2 HCl → H₂ + MgCl₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

First of all, you know that react 0.500 L of 2.00M HCl. Taking into account the definition of molarity. It is possible to calculate the number of moles that react as follows:

moles of HCl= 0.500 L× 2 M

moles of HCl= 1 mole

To calculate the number of moles H₂ formed, the following rule of three can be applied: if by reaction stoichiometry 2 moles of HCl form 1 mole of H₂, 1 mole of HCl form how many moles of H₂?

moles of H₂= (1 mole of HCl× 1 mole of H₂)÷ 2 moles of HCl

moles of H₂= 0.5 moles

Finally, to calculate the liters of H₂ formed, you can apply the following rule of three: if by definition of STP conditions 1 mole of H₂ occupies a volume of 22.4 liters, 0.5 moles occupies how much volume?

volume= (0.5 moles× 22.4 L)÷ 1 mole

volume= 11.2 L

In summary, 11.2 liters of hydrogen gas can be formed at STP.

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The reaction is a double displacement reaction, a precipitate of silver chloride, AgCl, is obtained, and the unknown compound is barium chloride, BaCl₂

Double displacement reactions are reactions that occur when two soluble ionic compounds are mixed together, and a precipitate is formed as a result of the exchange of radicals between the metallic ions in the ionic compounds.

An example of a double displacement reaction is the reaction between barium chloride and silver nitrate, where a precipitate of silver chloride is obtained. The equation of the reaction is given below:

BaCl₂ (aq) + AgNO₃ (aq) ---> Ba(NO₃)₂ (aq) + AgCl₂ (s)

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0.4g of Hydrogen will be produced when 4.6g of Sodium is used to provide Sodium Hydroxide.

The balanced equation for the formation of Sodium Hydroxide is given below:

2Na + 2H₂O ------> 2NaOH + H₂

It is shown that 2 moles of Na produce 2moles of Hydrogen. We know that the atomic mass of Sodium is 23g/mol. Then the number of moles present in 4.6g of Sodium would be,

Moles of Sodium = 4.6/23

Moles of Sodium = 0.2mol

Therefore,4.6g of Sodium has 0.2mol. Therefore, then 0.2 mol of sodium reacts to give out 0.2 mol of Hydrogen gas

The mass of hydrogen gas could be computed by the product of the mass of hydrogen gas and moles of hydrogen which is given by,

Mass of 2 atoms of hydrogen = 2x1x0.2

Mass of 2 atoms of hydrogen= 0.4g

Therefore, for every 0.2mol of Sodium, 0.1 mol of Hydrogen is produced which is equal to 0.4g of Hydrogen.

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The number of atoms of phosphorus that would be in 4.70 mol of copper (II) phosphate is 56.61 x [tex]10 ^{23[/tex] atoms.

According to an established rule, 1 mole of any substance contains one Avogadro's number, which is 6.022 x [tex]10 ^{23[/tex] atoms.

Copper (II) phosphate contains atoms of copper, phosphorus, and oxygen. In ion form, the compound ionizes according to the following equation:

[tex]Cu_3(PO_4)_2 --- > 3 Cu^2^+ + 2P^4^+ + 8O^2^-[/tex]

This means that every 1 mole of copper (II) phosphate contains 2 moles of phosphorus. The equivalent mole of phosphorus in 4.70 mol of copper (II) phosphate would be:

= 4.70 x 2

= 9.4 mol of phosphorus.

We said 1 mole of any substance =  6.022 x [tex]10 ^{23[/tex] atoms.

Hence, 9.4 moles of phosphorus will contain:

9.4 x  6.022 x [tex]10 ^{23[/tex] atoms

= 56.61 x [tex]10 ^{23[/tex] atoms

In other words, 4.70 mol of copper (II) phosphate will contain  56.61 x [tex]10 ^{23[/tex] atoms of phosphorus.

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As per the balanced reaction, 3 moles of calcium hydroxide produce 2 moles of aluminium hydroxide and 3 moles of calcium sulphate. The number of moles of calcium sulphate to produce 1.40 moles of aluminium hydroxide is 2.1 moles. Thus 2.1 moles of calcium sulphate will be produced.

Calcium sulphate is an inorganic compound formed from reaction of calcium hydroxide with any sulphates. The balanced equation of the given reaction is written as:

[tex]\rm Al_{2}(SO_{4})_{3} + 3Ca(OH)_{2} \rightarrow 2 Al(OH)_{3} + 3 CaSO_{4}[/tex]

As per the balanced reaction,  3 moles of calcium hydroxide  2 moles of aluminium hydroxide.Given that 1.40 moles of aluminium hydroxide is produced. The number of moles of calcium hydroxide required for these much product is:

= (1.40 × 3) 2 = 2.1.

3 moles of calcium hydroxide produce 3 moles of calcium sulphate. Thus, 2.1 moles gives 2.1 moles of product. Hence, number of moles of calcium sulphate produced is 2.1 moles .

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1.57 x 10²⁴ atoms of carbon to moles of carbon is 6.72 mole

Mole is the SI unit of amount of substance of a specified elementary entity

Here given data is

Atoms of carbon =  1.57 x 10²⁴

We have to calculate moles of carbon = ?

So one mole of carbon specifies 6.022×10²³ individual carbon atom that is mole of carbon

Then  1.57 × 10²⁴ atoms/6.022×10²³C×atom×mol⁻¹

Moles of carbon = 6.72 mole

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The number of grams of RbOH present in the solution can be found by using the formula

[tex]c = \frac{m}{M \times v} \\ [/tex]

where

m is the mass in grams

M is the molar mass in g/mol

v is the volume in L or dm³

making m the subject we have

m = c × M × V

First of all we have to find the molar mass of RbOH

M(RbOH) = 85.5 + 16 + 1 = 102.5 g/mol

v = 31 ml = 0.031 dm³ or 0.031 L

c = 5.6 M

Substituting it into the formula we have

m = 5.6 × 102.5 × 0.031 = 17.79 g

We have the final answer as

Hope this helps you

The redox reaction as it has been written can not be able to hold.

We know that a redox reaction is the kind of reaction that involves the oxidation and the reduction of a specie. In this case, we can see the reaction that occurs between the aluminum ion and sodium sulfide.

We know that that reaction can not be able to hold because the reduction potential of the aluminum ion is -1.66 V while the reduction potential of sodium is -2.71 V.

as such, sodium is more reducing than aluminum hence it can not be displaced by it.

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The concentration in parts per million of DDT are as follow:

3 x 10^-4 g in 2.3 kg minnow tissue PPM DDT is 0.13 ppm and in 717 µg in 1.5 kg breast milk is 0.478 ppm.

ppm DDT

PPM is defined as parts per million. It represents the lower concentration of the given solution. It also help in representing 0.001 gram or a milligram in the 1000 mL, which is equivalent to the 1 mg per liter.

Given,

A sample size of2.3 kg contained 3 x 10^-1 mg DDT

It means, 2.3 L sample contained 3 x 10^-1 mg DDT

Therefore, we can say that,

in part per million

= 3 x 10^-1 mg/ 2.3 L

= 0.13 mg/L

=0.13 ppm

A sample size of 1.5 kg contained 717 µg DDT

It means, 1.5 L sample contained 717 x 10^-3 mg DDT

Therefore, we can say that,

in part per million

= 717 x 10^-3 mg/ 1.5 L

= 0.478 mg/L

=0.478 ppm

Thus, we concluded that the concentration in parts per million of DDT are as follow:

3 x 10^-4 g in 2.3 kg minnow tissue PPM DDT is 0.13 ppm and in 717 µg in 1.5 kg breast milk is 0.478 ppm.

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1. The mole fraction of the ionic species KCl in the solution is 0.035

2. The molarity of the KCl solution is 1.908 M

3. The molality of the KCl solution is 2.027 M

1. How to determine the mole fraction of KCl

The mole fraction of KCl can be obtained as illustrated below:

Mole fraction of KCl = Mole of KCl / total mole

Mole fraction of KCl = 0.456 / 12.956

Mole fraction of KCl = 0.035

2. How to determine the molarity of the KCl solution

The molarity of the KCl solution can be obtained as follow:

Molarity = mole / volume

Molarity of KCl = 0.456 / 0.239

Molarity of KCl = 1.908 M

3. How to determine the molality of KCl solution

The molality of the KCl solution can be obtained as follow:

Molality = mole / mass of water (in Kg)

Molality of KCl = 0.456 / 0.225

Molality of KCl = 2.027 M

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The relationship that exists between smooth endoplasmic reticulum and cell membrane is that lipids are produced in the smooth endoplasmic reticulum which is the major product used for the synthesis of cell membrane.

The cell membrane is defined as part of the cell that is semi permeable which surrounds the cells and controls the passage of substances in and out of the cell.

The endoplasmic reticulum is a membrane bound organelle that are of two types namely:

The smooth endoplasmic reticulum is defined as the type of endoplasmic reticulum that is involved in the synthesis of lipids such as cholesterol and phospholipids.

A relationship exists between the cell membrane and the smooth endoplasmic reticulum because the lipids required for the synthesis of cell membrane are being produced by the smooth endoplasmic reticulum.

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Answer:

200 (kg / mol) x dL

Explanation:

The force of gravity needed to make a 13kg ball fall on Earth is 127.53N.

Force of gravity is a force that attracts any two objects with mass. The force of gravity is the weight of an object on Earth.

The weight of the force on an object due to the gravitational attraction between it and the Earth (or whatever astronomical object it is primarily influenced by).

The weight of an object can be calculated by multiplying the force of the object by acceleration due to gravity on Earth as follows:

Weight = force × 9.81m/s²

Weight = 13kg × 9.81m/s²

Weight = 127.53N.

Therefore, 127.53N is the weight of the ball on Earth.

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Patrick is a 16 year old boy whose body has stopped producing osteoclasts this means for his bones that his bones will become weak and brittle and other part of his body like the hip and wrist and spine will be affected by this

Osteoclast weaken brittle bones and making them prone to breaking from even minor stresses like coughing or stopping and the most frequent location for fracture caused by osteoclasts are the hip writ and spine and bone is a living tissue that constantly degrade and is replaced and by increasing their resorptive activity and destroying bone to initiate normal bone repair and the cell known as osteoclasts mediate bone loss in pathogenic circumstances

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According to the reaction illustrated, the amount, in moles, of copper (II) nitrate that will be formed is 0.456 mole.

The equation of the chemical reaction when silver nitrate reacts with copper is expressed as follows:

[tex]2AgNO_3 + Cu --- > Cu(NO_3)_2 +2 Ag[/tex]

The copper was in excess. This means that silver nitrate is the limiting reactant.

And 0.912 moles of the silver nitrate were reacted.

The mole ratio of silver nitrate and copper (II) nitrate according to the balanced equation of the reaction is 2:1. In other words, for every 1 mole of silver nitrate that reacts, 0.5 moles of copper (II) nitrate is formed.

Now, if 0.912 moles of silver nitrate is involved, the equivalent amount of moles of copper (II) nitrate that will be formed can be calculated as follows:

2 moles silver nitrate = 1 mole copper (II) nitrate

0.912 moles silver nitrate = 0.912 x 1/2

                                          = 0.456 mole

In other words, according to the reaction given, if 0.912 moles silver nitrate is reacted with excess copper, 0.456 mole copper (II) nitrate will be formed.

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