How many grams of Co, are produced when 88 g of o, is reacted with an excess of
butane?

Answer:NH3 + H2O <==> NH4+ + OH-

0.10 x 0.013 = 0.0013 M (this is the hydroxide concentration)

Kw = [H+] [H-]

1.00 x 10^-14 = (x) (0.0013)

x = 7.7 x 10^-12 M

Explanation:

Answer: The percentage of potassium iodide in the sample is 2.63 %.

Explanation:

The chemical equation for the reaction of iodide ions with bromine gas follows:

[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex]                (i)

The chemical equation for the reaction of iodate ions with barium ions follows:

[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex]                 ......(ii)

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of barium iodate = 0.0596 g

Molar mass of barium iodate = 487.13 g/mol

Using equation 1:

[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]

By Stoichiometry of the reaction (ii):

1 mole of barium iodate is produced by 2 moles of iodate ions

So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions

By the stoichiometry of the reaction (i):

1 mole of iodate ions are produced by 1 moles of iodine ions

So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions

Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]

Since, the molar mass of potassium iodide = 166 g/mol

Using equation 1:

[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]

To calculate the percentage by mass of a substance, the equation used is:

[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]

Mass of a solution = 1.54 g

Mass of potassium iodide = 0.0405 g

Using above equation:

[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]

Hence, the percentage of potassium iodide in the sample is 2.63 %.

Expressing the results of potassium iodide in percentage = 2.63%

The chemical reaction of iodine ions with Bromine gas can be expressed as :

I⁻  + 3Br₂  + 3H₂O  -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )

Chemical reaction between iodate ions with barium ions  can be expressed as :  Ba²⁺  + 2IO⁻₃ ------> Ba ( IO₃ )₂   --------- ( 2 )

Step 1 : Calculate the number of Barium iodate moles

mass of Barium iodate = 0.0596 g

molar mass of Barium iodate = 487.13 g/mol

from equation ( 1 )

moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles

also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions

= 2.44 * 10⁻⁴

molar mass of potassium iodide = 166 g/mol

Next step : Determine the mass of potassium iodide

moles of potassium * molar mass

= 2.44 * 10⁻⁴  * 166 g/mol  = 0.0405 g

Final step : Determine the percentage of potassium iodide in the solution

Percentage = ( mass of potassium iodide / mass of solution ) * 100

                   = ( 0.0405 / 1.54 ) * 100

                   = 2.63%

Hence we can conclude that potassium iodide in percentage = 2.63%

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Answer:

The answers to your questions are given below.

Explanation:

__ NaBr + CaF₂ —> __ NaF + CaBr₂

The above equation can be balance as follow:

NaBr + CaF₂ —> NaF + CaBr₂

There are 2 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 2 before NaF as shown below:

NaBr + CaF₂ —> 2NaF + CaBr₂

There are 2 atoms of Na on the right side and 1 atom on the left side. It can be balance by writing 2 before NaBr as shown below:

2NaBr + CaF₂ —> 2NaF + CaBr₂

Now, the equation is balanced.

Atom >>>> Reactant >>>> Product

Na >>>>>> 2 >>>>>>>>>>> 2

Br >>>>>>> 2 >>>>>>>>>>> 2

Ca >>>>>> 1 >>>>>>>>>>>> 1

F >>>>>>> 2 >>>>>>>>>>>> 2

......A) iconic bonds........

The attractive forces existing between the molecules of a substance are known as the intermolecular forces. Among the given forces dispersion bonds has the smallest energy per bond. Thus option D is correct.

The forces which are found in nonpolar molecules like O₂, N₂, Cl₂ etc. as well as in monoatomic gases such as 'He', 'Ne' etc. are called dispersion forces or London forces.

The London force is found to be the weakest intermolecular force and it is known to operate in all types of molecules. This force is also called the induced dipole - induced dipole attraction.

This force is generally resulted from the movement of electron cloud which creates a temporary instantaneous dipole. This distorts the arrangement of electrons of other atoms or molecules which are close to it and also induces a dipole in them.

In this way a large number of nonpolar molecules become polar temporarily and they are attracted by weak attractive forces. The dispersion bonds are easy to form and break.

The dispersion force is the weak attractive force and has the smallest energy per bond.

Thus the correct option is D - Dispersion bonds.

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Answer:

Helium and hydrogen......

Answer:

[tex]Kc=\frac{[HI]^2}{[I_2][H_2]}[/tex]

Explanation:

Hello there!

In this case, for these equilibrium problems it is firstly necessary to know the balanced reaction at equilibrium:

H2((g) + I2(g) ⇋ 2HI(g)

Next, by means of the law of mass action, it turns out possible for us to write the required and correct expression for the equilibrium constant by considering the concentrations and the coefficients in the aforementioned reaction:

[tex]Kc=\frac{[HI]^2}{[I_2][H_2]}[/tex]

Best regards!

Answer:

[tex]a=1.134\frac{m}{s^2}[/tex]

Explanation:

Hello there!

In this case, by considering the physical definition of force in terms of mass and acceleration:

[tex]F=m*a[/tex]

Given the generated force and the involved mass, we can compute the required acceleration as shown below:

[tex]a=\frac{F}{m}\\\\a=\frac{2552N}{114kg}[/tex]

Yet it is necessary to break out Newtons to:

[tex]a=\frac{2552\frac{kg*m}{s^2} }{114kg}\\\\a=1.134\frac{m}{s^2}[/tex]

Best regards!

Answer: 1. The empirical formula is [tex]PNF_2[/tex]  

2. The molecular formula is [tex]PNF_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P =[tex]\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles[/tex]

Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles[/tex]

Moles of F =[tex]\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = [tex]\frac{1.20}{1.20}=1[/tex]

For N = [tex]\frac{1.20}{1.20}=1[/tex]

For F =[tex]\frac{2.41}{1.20}=2[/tex]

The ratio of P: N: F= 1: 1: 2  

Hence the empirical formula is [tex]PNF_2[/tex]

The empirical weight of [tex]PNF_2[/tex]= 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1[/tex]

The molecular formula will be=[tex]1\times PNF_2=PNF_2[/tex]

Explanation:

c)The new kinetic energy of each gas particle is the sum of the kinetic energy that each particle is traveling with.

Answer:

Physical Change

Explanation:

Sublimation is a physical change. When a substance sublimes, it changes from a solid to a gas without going through the liquid phase.

Answer: [tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]

Explanation:

The skeletal thermochemical equation for the reaction is:

[tex]NH_3(g)+O_2(g)\rightarrow N_2O(g)+H_2O(l)+342kJ[/tex]

The balanced  thermochemical equation for the reaction is:

[tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]

When 1 mole of [tex]NH_3[/tex] reacts with oxygen , heat released = 342 kJ

Thus when 2 moles of [tex]NH_3[/tex] reacts with oxygen , heat released = [tex]\frac{2}{1}\times 342 kJ=684kJ[/tex]

Answer:

12

Explanation:

Boyles law

P1 (.36)

V1(28.3)

P2(.85)

V2(?)

P1*V1=P2*V2

plug it in, and you get 12

Answer:

c. uracil

Explanation:

uracil is not found in DNA. the missing base would be thymine

Answer:

80 ml

Explanation:

From the question,

Applying Dilution formular

MV = mv................... Equation 1

Where M = Molarity of Kl before dilution, V = Volume of Kl before dilution, m = molarity of Kl after dilution, v = volume of Kl after dilution.

make V the subject of the equation

V = mv/M............. Equation 2

Given: m = 1.00 M, v = 200 ml, M = 2.50 M

Substitute these values into equation 2

V = (1.00×200)/2.50

V = 80 ml

The concentration of SO42- ions in the final solution is 0.01 mol L-1.

The neutralisation reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) requires the use of a chemical equation that is balanced:

H2SO4 + NaOH = Na2SO4 + 2H2O

We may deduce from the balanced equation that one mole of H2SO4 reacts with one mole of Na2SO4, meaning that the amount of SO42- ions produced is equal to the amount of H2SO4 utilised.

We may determine how many moles of H2SO4 were used by calculating the amount of 0.05 M H2SO4 solution needed to neutralise 25 mL of NaOH solution:

H2SO4 moles are equal to the product of the volume of H2SO4 (in L) and the concentration of H2SO4 (in mol L-1): 0.03214 L 0.05 mol L-1 = 0.001607 mol

The formation of SO42- ions results in the same amount of moles as the
The resulting solution contains the following amounts of SO42- ions per mole of H2SO4 used:

Moles of SO42- ions/volume of solution (in L) = 0.001607 mol/(0.025 L + 0.03214 L) = 0.01 mol L-1; concentration of SO42- ions (in mol L-1)

As a result, the final solution contains 0.01 mol L-1 of SO42- ions.

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Answer: Intermolecular bonds are the forces between the molecules. Intramolecular are the forces within two atoms in a molecule.

Explanation:

Answer:

C. 1-ethyl, 3-methylcyclohexane

(Photo for proof at the bottom.)

Explanation:

The 1-ethyl is because you start numbering from the longest branch, towards the next closest branch. Prefix "eth-" means two, there are 2 carbons in the longest branch. 3-methyl is because the next branch is at number 3, and prefix "meth-" means 1, there is 1 carbon in that chain. "Cyclo" in cyclohexane means the skeletal model is shaped like a ring, and the "hexane" means there are 6 carbons in the ring. Prefix "hex" means 6.

Here's a photo of the unit review on Edge. Refer to the 2nd attachment for a visualization.

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Answer:

C. 1-ethyl,3-methylcyclohexane

Explanation:

Answer:

Scientists know that today the Earth's magnetic field is powered by the solidification of the planet's liquid iron core. The cooling and crystallization of the core stirs up the surrounding liquid iron, creating powerful electric currents that generate a magnetic field stretching far out into space.

Answer:

2nd option

Explanation:

Answer:

NO is the limiting reagent.  

In this reaction 0.886 mole of NO2 is produced

Explanation:

The chemical equation for this reaction is  

2NO(g) + O2(g) → 2NO2(g)

In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2  to produce 2 mole of 2NO2

0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2

0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2

Hence, NO is the limiting reagent.  

In this reaction 0.886 mole of NO2 is produced

Answer:

i think the answer is c.

Explanation:

Answer:

1455.6

Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values

35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6

Answer:

Explanation:

Elimination mechanism is the removal of two substituents  from a molecule in either a one- or two-step mechanism

Carbocation is a molecule containing a positive charged carbon atom and three bonds

The mechanism is generally accepted to always operate via an elimination step-wise process. The elimination mechanism process will always produce (after dehydration) a Zaitsev style alkene as major product. The driving force for the production of this Zaitsev style alkene product is generally going to be determined by stability of the carbocation.

Dehydration is a process in which a water molecule releases from the given compound to form an unsaturated compound.

Dehydration process of alcohol is done by the elimination reaction in which water molecule eliminate and due to this dehydration we get an unsaturated product alkene. This dehydration process is go through the formation of carbocation which in turn forms the product. Stability of the product is determined by the Zaitsev rule and according to this rule most substituted product is the major and stable product.

Hence correct options are: Elimination, elimination, Zaitsev, Zaitsev and carbocation.

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Answer:

The real volume of the gas particles is extremely small when compared to the entire volume of their container.

d)Gas particles constantly expand and contract, which constantly alters their volume. When these volumes are averaged, a value of zero is obtained

Explanation: