Draw clearly the structural formulas of the given compounds. (a) 3-methyl heptane​

Answer:

44.8 grams

Explanation:

The density exists as a ratio comparing the mass of a piece of copper per 1 cm³. Since you have been given a new volume, you can set up a proportion to find the new mass.

[tex]\frac{8.96 grams}{1.00 cm^3} =\frac{?grams}{5.00cm^3}[/tex]                                         <----- Proportion

[tex]44.8 = (1.00cm^3)(?grams)[/tex]                              <----- Cross-multiply

[tex]44.8= ?grams[/tex]                                                 <----- Divide both sides by 1.00

Answer:

.512 ATM

Explanation:

P1V1 = P2V2

2 * 5   =  P2 * 19.5

P2 = .512 ATM

Answer:

1.2   ( two significant digits)

Explanation:

5.4 - 4.21        answer should only have the the number of significant digits as the number in the equation with the least number of significant digits... in this case   5.4   is only two significant digits....so the answer should only have two significant digits

5.4 - 4.21 = 1.19     round to two digits = 1.2    

pH = pKa + log [conj.base]/[acid]

Let the concentration of the conjugate base = x

=> The concentration of the acid is also x (As the acid concentration is equivalent to that of the conjugate base)

for CH₃COOH=0.50 M and CH₃COO=0,25 M

pH = pKa + log [salt]/[acid]

pKa = -log Ka = -log 1.8x10-5 = 4.74

pH = 4.74 + log (0.50 M/1.0 M) = 4.74 + log 0.5 = 4.74 + (-0.3)

pH = 4.74 - 0.3

pH = 4.44

For0.20 M Na2CO3

pH=−log[H+]

    =−log (1.54×10−12)

   =11.8

for HNO₂

Kb = 2.17X10^-11 = x^2 / 0.148

x = [OH-] = 1.79X10^-6

pOH = 5.747

pH = 14.000 - pOH = 8.253

​

for HPO₄³⁻

= [HPO₄³⁻] [PO₄³⁻] [H +]/×10⁻³

=[PO₄³⁻]5.464×10⁻¹⁸

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The change to the experimental design that would improve the reliability of the engineers' measurements is C. testing single samples from more than three locations

It should be noted that experimental design is the process of carrying out research in an objective as well as a controlled fashion in order for precision to be maximized and specific conclusions can be drawn from the hypothesis statement.

Based on the information, the locations selected are only three. Also, an area must have more types of soil with variable porosity. Hence, the collection of more soil samples may provide more accurate results than collecting soil samples from only three regions.

Therefore, change to the experimental design that would improve the reliability of the engineers' measurements is the testing single samples from more than three locations.

In conclusion, the correct option is C.

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Complete question:

Which change to the experimental design would improve the reliability of the engineers' measurements?

A. using a liquid other than water to determine porosity

B. using flasks instead of beakers

C. testing single samples from more than three locations

D. testing more samples from each location

The two compounds that are the correct enol tautomers of 2-butnone are the structures shown in Option B.

Isomerism is a situation in which two compounds have the same molecular formula but do not have the same structural formula. This occurs because the atoms in the molecule are arranged differently in space. One of the types of isomerism that we have is the Keto Enol Tautomerism. The Keto Enol Tautomerism occurs between the Keto and the enol forms of a compound via the movement of electrons. It is common in ketone compounds.

In this case, we can see that the compound 2-butanone can give rise to two enol compounds that have the same molecular formula as the 2-butanone.

The two compounds that are the correct enol tautomers of 2-butnone are the structures shown in Option B.

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The common word on the outside of a shipping container that will alert you to the fact that the material inside is hazardous is caution.

Hazardous statements are those words or phrases that are labelled against chemicals which shows the degree of how dangerous the chemical are.

The common words that are used as hazardous statements which shows that a chemical is dangerous include the following:

Therefore, the common word on the outside of a shipping container that will alert you to the fact that the material inside is hazardous is caution.

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S denotes the ''number of shared electron pairs by an atom'', Hence, Silicon transmits its 4 electrons with 4 Cl-atoms. Thus the Value of S is 4 in SiCl₄ .It is also known as covalency

Silicon Tetrachloride is a transparent, colorless, fuming liquid with an irritating odor. It is used to create Silicon-containing chemicals and in chemical analysis. determine potentially hazardous frontages.

Silicon tetrachloride is designed by the chlorination of various silicon compounds such as ferrosilicon, silicon carbide, or mixtures of silicon dioxide and carbon. The ferrosilicon route is most familiar. In the laboratory, SiCl 4 can be prepared by dining silicon with chlorine on 600 °C (1,112 °F): Si + 2 Cl 2 → SiCl.

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The oxidation number of reactant Fe is 0 while the  oxidation number of iron in the product is +3

The term redox reaction implies a reaction in which there is an increase in the oxidation number of a specie and the decrease in the oxidation number of another specie.

Now we have the answers as follows;

1) The oxidation number of reactant Fe is 0

2) The oxidation number of reactant oxygen is 0

3) The oxidation number of iron in the product is +3

4) The oxidation number of oxygen in the product is -2

5). Iron is oxidized in the reaction

6) Oxygen is reduced in the reaction

7) The oxidizing agent in this case is the oxygen atom

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The equilibrium shift if the HCl concentration is increased to the product.

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

If the concentration of a substance is changed, then the equilibrium will shift in such a way that it minimise the effect of change which occur. If we increase the concentration of a reactant, then the equilibrium will shift to minimize the changes in the direction of the reaction which uses the reactants, so that the reactant concentration decreases.

Thus, we concluded that with increase in the concentration of reactant equilibrium shift forward.

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Mass of [tex]$\mathrm{Na}{2} \mathrm{CO} 3$[/tex] required [tex]$=0.55 \mathrm{~g}$[/tex]

The mass 0.120M HCI:

Moles of [tex]$\mathrm{HCl}$[/tex]reacted [tex]$=0.120 \times 21.6 / 1000=0.00260$[/tex]

according to balanced reaction, [tex]$\mathrm{HCl}$ and $\mathrm{Na} 2 \mathrm{CO} 3$[/tex] reacts in [tex]$2: 1$[/tex] the ratio

moles of [tex]$\mathrm{Na} 2 \mathrm{CO} 3$[/tex]required [tex]$=0.00260 \times 2=0.00520$[/tex]

convert moles to mass

mass [tex]$=$[/tex]moles [tex]$\times$[/tex]molar mass

mass [tex]$=0.00520 \times 106.0$[/tex]

mass [tex]$=0.55 \mathrm{~g}$[/tex]

mass of [tex]$\mathrm{Na} 2 \mathrm{CO} 3$[/tex]required [tex]$=0.55 \mathrm{~g}$[/tex]

Mass 0.120M HC is [tex]$=0.55 \mathrm{~g}$[/tex]

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Answer:

Molar mass = 30.026 g/mol

Explanation:

Let Molar mass be mm

mm of CH2O = 12.01 + 2(1.008) + 16.00

= 12.01 + 2.016 + 16.00 = 30.026 g/mol

Answer:

C. zero 0

Explanation:

single elements will always possess an oxication number of zero

As we know that the equivalency point of moles of added base will be equal to the moles of original acid, this allows to determine the number of moles of original acid.

At the equivalency point the moles of added base will be equal to the moles of original acid, this allows the determination of the number of moles of original acid. This can be combined with the original volume of the analyte solution in order to determine its concentration.

So we can conclude that at the equivalence point the moles of added base will be equal to the moles of original acid, this allows the determination of the number of moles of original acid.

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Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.

In first place, the balanced reaction is:

Cu + 2 AgNO₃  → 2 Ag + Cu(NO₃)₂

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:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?

[tex]amount of moles of AgNO_{3} =\frac{1.8 moles of Cux2 moles of AgNO_{3} }{1 mole of Cu}[/tex]

amount of moles of AgNO₃= 3.6 moles

But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.

In summary, AgNO₃ will be the limiting reagent.

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Answer:

445 mL NH₄Br

Explanation:

(Step 1)

Convert grams ammonium bromide (NH₄Br) to moles using its molar mass.

Molar Mass (NH₄Br): 14.007 g/mol + 4(1.008 g/mol) + 79.904 g/mol

Molar Mass (NH₄Br): 97.943 g/mol

8.42 grams NH₄Br                1 mole
-----------------------------  x  ------------------------  =  0.0860 moles NH₄Br
                                         97.943 grams

(Step 2)

Calculate the volume of NH₄Br using the molarity equation. Then, convert liters (L) to milliliters (mL).

Molarity (M) = moles / volume (L)

0.193 M = 0.0860 moles / volume

(0.193 M) x volume = 0.0860 moles

volume = 0.445 L

1,000 mL = 1 L

0.445 L NH₄Br           1,000 mL
------------------------  x  ------------------  =  445 mL NH₄Br
                                        1 L

Answer:

okkkk zjjdjdjdjduxuxu

ClO(4)− is regarded as perchlorate ion and are produced commercially in most situations as salts via industries and in the laboratory.

This ion is referred to as a monovalent inorganic anion and is obtained by deprotonation of perchloric acid. It is composed of chlorine and oxygen atoms in the ratio 1 to 4 respectively.

This has  32 valence electrons available in the Lewis structure and is used in the commercial production of solid rocket fuel.This ion has a molar mass of 99.451 g mol−1 and is used in different processes such as an oxidizer and to control static electricity during the process of food preservation in industries.

Therefore ClO₄− is also regarded as perchlorate ion and is the most appropriate choice.

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Answer:0

Explanation:

Since oxygen gas i.e., diatomic oxygen exists in the atmosphere in the free state, hence the oxidation state of each atom will be 0

Answer:

See below

Explanation:

370 g  /  .090 g/l = _____?___   liters

take that answer    then :

  ? liters   /   22.4 liters / mole   = # moles     done!

There are approximately 183.50 moles of hydrogen gas in 370.g of hydrogen gas.

The number of moles in 370.g of hydrogen gas using the given information.

In the given example,

Density of hydrogen gas = 0.090 g/L

Molar volume at normal pressure and 0.214°C = 22.4 L/mol

So, we  calculate the volume of hydrogen gas (in L) using the given mass and density:

Volume (in L) = mass (in g) / density (in g/L)

Volume (in L) = 370.g / 0.090 g/L

Volume (in L) ≈ 4111.11 L

By using the molar volume, calculate the number of moles:

moles = Volume (in L) / Molar volume (in L/mol)

moles = 4111.11 L / 22.4 L/mol

moles ≈ 183.50 moles (rounded to two decimal places)

Hence, there are approximately 183.50 moles of hydrogen gas in 370.g of hydrogen gas.

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Answer:

Explanation:

Discussion

When Pressure increases  equilibrium shifts to the side with the smallest number of moles. But which side is that?

N2(g) + 3H2(g) ⇌ 2NH3(g)

The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols

on the left side.

The right side has 2 mols of NH3 = 2 mols on the right.

Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.

Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.

2NH3 is only two mols.

The equilibrium shifts Right

Answer

D

Answer:

2.52 g NaCl

Explanation:

(Step 1)

To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:

Molarity = moles / volume (L)

After you convert mL to L, you can plug the given values into the equation and simplify to find moles.

136.9 mL / 1,000 = 0.1369 L

Molarity = moles / volume

0.315 M = moles / 0.1369 L

0.0431 = moles

(Step 2)

Now, you can use the molar mass to convert moles to grams.

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

0.0431 moles NaCl           58.443 g
------------------------------  x  -------------------  =  2.52 g NaCl
                                            1 mole

Answer:

X = ¹₀n

Explanation:

The mass number is the total amount of neutrons and protons in the element. The atomic number is the total protons in the element. The mass number is the superscript and the atomic number is the subscript. The mass and atomic numbers must be balanced on both sides of the equation. The element can be identified by the atomic number.

In the event that we have an atomic number of zero, there is no element this represents, so the letter "n" should be used.

²³₁₁Na + ¹₁H  →  ²³₁₂Mg + X

Mass Number:

Reactants = Products

23 + 1 = 23 + ?

24 = 23 + ?

1 = ?

Atomic Number:

Reactants = Products

11 + 1 = 12 + ?

12 = 12 + ?

0 = ?

The complete equation is:

²³₁₁Na + ¹₁H  →  ²³₁₂Mg + ¹₀n

Answer:

The answer would be D.

Explanation:

Whenever one has many of the same correlating numbers within in a pattern, the rest tend to follow. With the information I was given and not the actual plotting, that would be my answer.

Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.

In a reaction, we can judge using the value of the equilibrium constant weather or not the reaction moves on to completion. If the reaction moves up to completion, it the follows that the value of the equilibrium constant ought to be large.

On the other hand, when we have a case that the equilibrium constant is small and is not so large, then the reaction does not proceed essentially to completion.

Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.

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Answer:

[tex]\boxed{\sf no.\ of \ molecules = 7.13 \times 10^{22} \ molecules}[/tex]

Explanation:

Mass in g = m = 5.68 g

Molar mass = M = (16)3 = 48 g/mol

Avogadro's number = [tex]N_A[/tex] = 6.023 × 10²³

No. of molecules = ?

[tex]\displaystyle no.\ of \ molecules = \frac{m}{M} \times N_A[/tex]

Put the givens in the above formula

[tex]\displaystyle no. \ of \ molecules =\frac{5.68}{48} \times 6.023 \times 10^{23}\\\\no. \ of\ molecules = 0.12 \times 6.023 \times 10^{23}\\\\no. \ of \ molecules = 0.713\times 10^{23}\\\\no.\ of \ molecules = 7.13 \times 10^{22} \ molecules\\\\\rule[225]{225}{2}[/tex]

The oxidation number of chlorine in KClO3 is +5 (option A).

The oxidation number of an element is the hypothetical charge of an atom within a molecule.

The oxidation number of an element like chlorine in a compound like Pottasium chlorate can be calculated as follows:

The oxidation number of the elements in KClO3 is as follows:

1 + x - 2(3) = 0

x - 5 = 0

x = +5

Therefore, the oxidation number of chlorine in KClO3 is +5.

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The concentration of carbon monoxide at equilibrium is 0.209 M.

CO2 = 2 mol/5 L = 0.4

H2 = 1.5 mol/5L = 0.3

Create ICE table as shown below

        CO2(g)    +     H2(g) ↔       CO(g)    +     H2O(g)

I          0.4               0.3                0                  0

C        - x                 - x                 x                   x

E        0.4 - x           0.3 - x           x                   x

Kc = [CO][H₂O] / [CO₂][H₂]

2.5 = (x²)/(0.4 - x)(0.3 - x)

x² = 2.5(0.4 - x)(0.3 - x)

x² = 2.5(0.12 - 0.7x + x²)

x²  = 0.3 - 1.75x + 2.5x²

0 = 1.5x² - 1.75x + 0.3

solve the quadratic equation using formula method;

x = 0.209

Thus, the concentration of carbon monoxide at equilibrium is 0.209 M.

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Answer:

You can never have above 100% yield under any circumstance

To solve such this we must know the concept of chemical reaction. Therefore, the correct option is option D that is  you can never have above 100% yield under any circumstance.

Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.

You can never have above 100% yield under any circumstance for any chemical reactions. Option D is the correct options among all the given options.

Therefore, the correct option is option D that is  you can never have above 100% yield under any circumstance.

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