Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H2O that would form when 2.78 mol NH3 and 5.19 mol O2 react.

Answer:

Excluding any secondary chemical reactions,

which would be more effective as an antifreeze; a

solution containing

Select one:

25m CH3OH

the combination of 25m CH3OH and 25m

KCI.

25m KCI.

Explanation:

Antifreeze is the one that reduces the freezing point of a solvent further and will not allow the solvent to freeze.

Among the given options, the correct option is:

25 m CH3OH and 25m KCl.

Since, KOH is a strong electrolyte and dissociates into two ions.

So, the freezing point of solvent decreases further.

Answer:

The molar mass of N2O5 is 108 g/mol. 1296g of N2O5 has 1296 / 108 = 12 moles. 1 mole contains 6.022 x 10^23 molecules

Answer:

Explanation:

In a pure metal, the electrons can be thought of as [concentrated] around atoms throughout the metal. Using molecular orbital theory, there [is ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [antibonding] orbitals are typically higher in energy and are mostly (filled]

Answer:

It predicts whether or not a reaction will be spontaneous.

Explanation:

The equation;

∆G= ∆H - T∆S enables us to obtain the Gibbs free energy of a chemical process.

The Gibbs free energy value tells us whether a chemical process will be spontaneous or not.

When;

∆G>0 the reaction is not spontaneous

∆G=0 the reaction is at equilibrium

∆G<0 the reaction is spontaneous

Answer:

C₆H₆

Explanation:

Each border of the figure represents 1 atom of carbon. We have 6 borders = 6 atoms of carbon.

Each atom of carbon form 4 bonds. All the carbons are doing a double bond and a single bond with other carbons. That means are bonded 3 times. The other bond (That is not represented in the figure. See the image) comes from hydrogens. As we have 6 carbons that are bonded each 1 with one hydrogen. There are six hydrogens and the molecular formula is:

This structure is: Benzene

Answer:

Mass of sodium produced = 2.74 × 10⁵ g of Na

Mass of chlorine produce = 4.24 × 10⁵ g of Cl₂

Explanation:

In the electrolysis of molten NaCl as described above, the quantity of charge used is given by the formula, Q = I × t

Where I isnthe current passed in amperes and t is time in seconds.

Q = 4.0 × 10⁴ A × (8 × 60 × 60) s = 1.152 × 10⁹ C

Equation for the discharge of sodium is; Na+ + e- ---> Na (s)

One mole of electrons is required to discharge one mole of Na

One mole of electron = 1 faraday = 96500 C

One mole of Na has a mass of 23 g

96500 C produces 23 g of Na

1.152 × 10⁹ C will produce 23 g × 1.152 × 10⁹ C / 96500 C = 2.74 × 10⁵ g of Na

Equation for the discharge of chlorine gas is; 2 Cl- ---> Cl₂(g) + 2e-

Two mole of electrons are required to discharge one mole of chlorine gas

Two moles of electron = 2 faraday = 2 × 96500 C = 193000

One mole of Cl₂ has a mass of 71 g

193000 C produces 71 g of Cl₂

1.152 × 10⁹ C will produce 71 g × 1.152 × 10⁹ C / 193000 C = 4.24 × 10⁵ g of Cl₂

The amount of Na produced is 274551 g and the amount of Cl₂ produced is 423763.5 g.

Current passed through the cell = [tex]4.0\times10^4 A[/tex]

Time = 8 Hours

We have to calculate the amount of Na and [tex]Cl_2[/tex] produced in 8 hours in the downs cell.

The Downs process is an electrochemical method for the commercial preparation of metallic sodium, in which molten NaCl is electrolyzed in a special apparatus called the Downs cell.

The total charge passed through the cell is calculated by the given formula as

Charge(Q) = Current(I) × time(t)

Q = [tex]4.0\times10^4 A \times t[/tex]

[tex]t = 8 \times 60\times 60[/tex] sec

t = 28800 sec

Q = [tex]4.0\times10^4 A \times 28800[/tex] sec

Q = [tex]115200\times 10^4 A\ sec[/tex]

We know that, Ampere = Coulombs per sec

Q = [tex]115200\times 10^4\ C[/tex]

1 mol of electrons  96500 C charge

Therefore, the number of mols of electrons carries this [tex]115200\times 10^4\ C[/tex] charge = [tex]\frac{115200\times10^4}{96500}[/tex] = 1.1937 × 10⁴ = 11937 mol electrons

In the Down's cell

Half cell reactions are:

Reduction half-reaction: [tex]2Na^+(aq)+2e^- \to 2Na(s)[/tex]

Oxidation half-reaction: [tex]2Cl^-(aq)\to Cl_2(g) + 2e^-[/tex]

We know that no. of moles = [tex]\frac{given \ mass}{molar \ mass}[/tex]

Molar mass of Na = 23 g/mol

The mass of Na formed = 11937 mol × 23 g/mol = 274551 grams

The molar mass of Cl₂ = 71 g/mol

The mass of Cl₂ = [tex]\frac{11937}{2}\times 71 = 423763.5 \ grams[/tex]

Hence, the amount of Na produced is 274551 g and the amount of Cl₂ produced is 423763.5 g.

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Answer:

Earth has a mass of 5.9736×1024 kg

hope it is helpful to you

Answer:

H2

Explanation:

Answer:

[tex]{ \bf{H _{2} }} \\ { \tt{hydrogen \: gas}}[/tex]

Answer:

1.6 x 10⁻²⁴ moles

Explanation:

We have 0.984 molecules of F₂. We know that 1 mol is equal to 6.022 x 10²³ molecules. Thus, we have a conversion factor: 1 mol/6.022 x 10²³ molecules

So, we multiply the molecules of F₂ by the conversion factor to calculate the moles:

0.984 molecules x  1 mol/6.022 x 10²³ molecules = 1.6 x 10⁻²⁴ moles

Answer:

10.5 × 10^5 M

Explanation:

E°cell = E°cathode - E°anode

E°cell = 0.85 - (-0.74) = 1.59 V

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

1.499 = 1.59 - 0.0592/6 log [Cr^3+]/6.35x10^-4

1.499 - 1.59 = - 0.0592/6 log [Cr^3+]/6.35x10^-4

-0.091 = -0.00987 log [Cr^3+]/6.35x10^-4

-0.091/ -0.00987 = log [Cr^3+]/6.35x10^-4

9.22 = log [Cr^3+]/6.35x10^-4

Antilog (9.22) = [Cr^3+]/6.35x10^-4

1.66 × 10^9 = [Cr^3+]/6.35x10^-4

[Cr^3+] = 1.66 × 10^9 × 6.35x10^-4

[Cr^3+] = 10.5 × 10^5 M

Answer:

Li(s) + H2O(l) -----> LiOH(aq) + H2(g)

Ca(s) + 2H2O(l) ---------------> Ca(OH)2(aq) + H2(g)

Explanation:

Metals react with water to yield the metal hydroxide and hydrogen gas as shown in the answer above.

The reaction equations were balanced, the number of atoms of each element on both side of the reaction equation is exactly the same.

This is the way to write a balanced reaction equation for the species shown in the question.

Answer:

Magnesium.

Explanation:

Because it is in group II

Answer:

2

Explanation:

2 alkenes are formed by E2 elimination of HBr from 3-bromo -3, 4 - dimethylhexane using a strong base such as sodium methoxide

Answer:

energy is the ability to do work

Answer:

See explanation

Explanation:

First we divide the percentage by mass of each element by it's relative atomic mass then we divide the quotients obtained by the lowest ratio obtained in the first step.

C- 76.57/12, H= 6.43/1, O = 17.00/16

C- 6.38/1.06, H= 6.43/1.06, O= 1.06/1.06

C- 6, H- 6, O- 1

Empirical formula: C6H6O

[(12 ×6) + (6 × 1) + (16 × 1)]n=94.11

[72 + 6 +16]n = 94.11

n = 94.11/94

n= 1

Molecular formula = C6H6O

2)

C- 53.30/12, H- 11.19/1, O- 35.51/16

C- 4.44/2.22, H- 11.19/2.22, O- 2.22/2.22

C- 2, H- 5, O- 1

Empirical formula: C2H5O

[(2×12) + (5× 1) + (1×16)]n = 90.12

[24 + 5 + 16] n = 90.12

n= 90.12/45

n= 2

Molecular formula = C4H10O2

Answer:

[tex]n=6\ to\ n=3[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=274 *10^{-9}m[/tex]

Bohr's constant [tex]R = 1.097 × 10^7 / m (or m−1)[/tex]

Helium atom [tex]z=2[/tex]

Generally the equation for Wavelength is mathematically given by

 [tex]\frac{1}{\lambda}=Rz^2(\frac{1}{nf^2}-\frac{1}{nf^2})[/tex]

 [tex]0.083=(\frac{1}{nf^2}-\frac{1}{nf^2})[/tex]

Therefore

The Range of n fall at

 [tex]n=6\ to\ n=3[/tex]

Answer:

2.29 M

Explanation:

Equation of the reaction;

Ca(OH)2(aq) + 2HClO4(aq) → 2H2O(l) + Ca(ClO4)2(aq)

Concentration of acid CA = 1.49 M

Concentration of base CB= ????

Volume of acid VA= 43.0 ml

Volume of base VB= 14.0 ml

Number of moles of acid NA = 2 moles

Number of moles of base NB = 1 mole

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

CB= CAVANB/VBNA

CB= 1.49 × 43.0 × 1/14.0 × 2

CB= 2.29 M

A is the answer

In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place

dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the

outside and doing the central atom last if there are enough. Do not exceed the total number of valence electrons

identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate

the presence of two valence electrons

Answer:

Explanation: i did it

Answer:

txyxyc8uviciycuyc

Explanation:

gghvj ucy7cyvyfy

Answer: When excited sodium atoms may emit radiation having a wavelength of 589nm. It's wavelength in meters is [tex]589 \times 10^{-9} m[/tex].

Explanation:

Given: Wavelength = 589 nm

It is known that,

[tex]1 nm = 10^{-9} m[/tex]

Hence, 589 nm is converted into meters as follows.

[tex]589 nm = 589 nm \times \frac{10^{-9}m}{1 nm}\\= 589 \times 10^{-9} m[/tex]

Thus, we can conclude that when excited sodium atoms may emit radiation having a wavelength of 589nm. It's wavelength in meters is [tex]589 \times 10^{-9} m[/tex].

Answer:

b. glass and charcoal

Explanation:

Step 1: Given data

Step 2: Determine which material will float in molten lead

Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.

Answer: In order for a substitution reaction to take place, at a minimum, loss of a leaving group and nucleophilic attack are needed.

Explanation:

When one group in a molecule is replaced by another one then it is called a substitution reaction.

For a substitution reaction reaction to occur at a minimum, it is necessary to have a leaving group that helps in creating a carbocation and hence a nucleophile can easily attack.

For example, [tex]CH_{4} + Cl_{2} + h\nu \rightarrow CH_{3}Cl + HCl[/tex]

Thus, we can conclude that in order for a substitution reaction to take place, at a minimum, loss of a leaving group and nucleophilic attack are needed.

Answer: The balanced half-equations for silver + oxygen= silver oxide are:

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

Explanation:

The word equation is as follows.

silver + oxygen = silver oxide

In terms of chemical formulas this equation can be written as follows.

[tex]Ag + O_{2} \rightarrow Ag_{2}O[/tex]

The removal on electron(s) from an atom, ion or molecule in a chemical reaction is called oxidation.

The gain of electron(s) by an atom, ion or molecule in a chemical reaction is called reduction.

Hence, half-reaction equations for the given reaction is as follows.

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

As the number of atoms participating in the reaction are equal. Hence, the half-equations are balanced.

Thus, we can conclude that the balanced half-equations for silver + oxygen = silver oxide are:

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

Answer:

See explanation

Explanation:

Equation of the reaction;

Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)

Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3

Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl

1 mole of Na2CO3 reacts with 2 moles of HCl

0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl

Hence Na2CO3 is the limiting reactant

Since there is 0.6 moles of HCl present, the number of moles of excess reagent=

0.6 moles - 0.4 moles = 0.2 moles of HCl

1 mole of Na2CO3 forms 1 mole of water

0.2 moles of Na2CO3 forms 0.2 moles of water

Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water

1 mole of Na2CO3 yields 1 mole of CO2

0.2 moles of Na2CO3 yields 0.2 moles of CO2

1 mole of CO2 occupies 22.4 L

0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP

Hence;

V1=4.48 L

T1 = 273 K

P1= 760 mmHg

T2 = 27°C + 273 = 300 K

P2 = 760 mmHg

V2 =

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 × 4.48 × 300/760 × 273

V2= 4.9 L

The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.

hope it helps to everyone

Answer:

The general preparation of esters( for example ethyl ethanoate) is through a process known as ESTERIFICATION.

Explanation:

The formation of an ester by the reaction between an alkanol and an acid is known as esterification. This reaction is extremely slow and reversible at room temperature, and is catalyzed by a high concentration of hydrogen ions.

In the preparation of one of the simpler esters known as ETHYL ETHANOATE the reactants include ethanol(an alcohol) and glacial ethanoic acid(a carboxylic acid) in the presence of concentrated tetraoxosulphate VI acid as a CATALYST. Note that, a catalyst is any substance that is able to increase the rate of a chemical reaction.

The mixture is warmed in a water bath( hot but not boiling) for about 25 minutes. The mixture is poured into a beaker partially filled with a sodium or calcium chloride to remove interacted ethanol. The ethyl ETHANOATE floats on the mixture as oily globules.

Answer:

5.0molO

Explanation:

To find the moles of oxygen in 2.5 moles of caffeine, we will first research caffeine's molecular formula: C8H10N4O2. From the molecular formula, we can see there are 2 oxygen atoms in every 1 molecule of C8H10N4O2.We can therefore multiply by the following mole ratio to get the moles of oxygen.

2.5molC8H10N4O2×2molO/1molC8H10N4O2 = 5.0molO.

Answer:

e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature

Answer:

D. Temperature and activation energy is the correct answer

Explanation:

^_^

molarity = moles of solute / volume of solution

moles of solute = molarity × volume of solution

moles of solute = 0.245 mol/L × 0.1500 L

moles of solute = 0.03675 mol

moles of solute = 0.0368 mol

-----------------------------------------------------------

Step 1: Calculate the molar mass of solute.

molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)

molar mass of solute = 110.98 g/mol

Step 2: Calculate the mass of solute.

mass of solute = moles of solute × molar mass of solute

mass of solute = 0.03675 mol × 110.98 g/mol

mass of solute = 4.08 g

Note: The volume of solution must be expressed in liters (L).

Answer:

[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]

[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]

In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.

First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.

Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.

[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]

We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]

The units of liters cancel.

[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]

[tex]0.03675 \ mol \ CaCl_2[/tex]

The original measurements have 3 significant figures, so our answer must have the same.

We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.

[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]

We can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.

The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.

Notice that chlorine has a subscript of 2. We must multiply the molar mass by 2.

Add calcium's molar mass.

Use the molar mass as a ratio.

[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

Multiply the moles of calcium chloride we calculated above.

[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

The units of moles of calcium chloride cancel.

[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]

[tex]4.084064 \ g\ CaCl_2[/tex]

Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.

[tex]\bold {4.08 \ g \ CaCl_2}[/tex]