When naming organic compounds, there are strict rules regarding punctuation.
1. A comma is used to separate two numbers.
2. A hyphen is used to separate a number from a letter.
Rewrite the name of this compound using hyphens and commas as appropriate.

Answer:

Age of rock = 6.12 × 10³ years

Note: The question is incomplete.A similar but complete question is given below.

The half-life for the radioactive decay of carbon-14 to nitrogen-14 is 5.73 x 10^3 years. Suppose nuclear chemical analysis shows that there is 0.523mmol of nitrogen-14 for every 1.000 mmol of carbon-14 in a certain sample of rock.

Calculate the age of the rock. Round your answer to 2 significant digits.

Explanation:

The half-life of a radioactive material is the time taken for half the atoms in the atomic nucleus of a material to disintegrate.

The half-life for the radioactive decay of carbon-14 to nitrogen-14 is given as 5.73 x 10³ years. This means that given 1 mole of carbon-14 is present initially, after one half-life, 0.5 moles of carbon-14 would remain.

Number of millimoles of carbon-14 remaining = 1 - 0.523 = 0.477 mmol

Number of half-lives that the carbon-14 has undergone is determined as follows:

Amount remaining = (1/2)ⁿ

where nnis number of half-lives

0.5 mmol = one half-life

0.5 = (1/2)¹

O.477 = (1/2)ⁿ = (0.5)ⁿ

㏒₀.₅(0.477) = n

n = ㏒(0.477)/㏒(0.5)

n = 1.067938829

Age of the rock = number of half-lives × half-life

Age of rock = 1.067938829 × 5.73 × 10³ years

Age of rock = 6.12 × 10³ years

Answer:

18.82 mg

Explanation:

From the given information:

The molar mass of CO2  is calculated as follow

= (12 + (16 ×2))

= 44

The mass of carbon is determined by dividing the mass no of carbon from co2 by the molar mass of CO2,  followed by multiplying it by 69.00 mg

= [tex](\dfrac{12}{44}\times 69 )[/tex]

=(0.2727 × 69 )

= 18.82 mg

Answer:

The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Explanation:

The structure of nucleic acid polymers is built up from monomers of nucleotides.

A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.

When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.

Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Answer:

pH = 10.20

Explanation:

The HClO reacts with NaOH as follows:

HClO + NaOH → H2O + NaClO

Where HClO and NaOH react in a 1:1 reaction.

As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.

And the NaClO produced decreases its concentration in 2 because the volume is doubled.

The concentration of NaClO is: 0.150M / 2 = 0.075M

The equilibrium of NaClO is:

NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)

Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =

1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]

[NaClO] = 0.075M

As both HClO and OH- comes from the same equilibrium,

[HClO] = [OH-] = X

Where X is the reactoin coordinate

Replacing:

3.33x10⁻⁷ = [X] [X] / [0.075M]

2.5x10⁻⁸ = X²

X = 1.58x10⁻⁴M = [OH-]

pOH = -log [OH-]

pOH = 3.80

pH = 14 - pOH

the rate would decrease because the reactants are being depleted.

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

Answer:

Solids are made up of cation-anion pairs, but plasmas are not

Explanation:

Answer:

3,29L

Explanation:

3.29L = V2

Formula: V1/T1 = V2/T2

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

Given:

V1 = 3.0 L V2 = ?

T1 = 310 K T2 = 340 K

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

Plugin:

(X stands in place of V2 just to make it easier to look at)

[3.0L / 310K = X / 340K]

(3.0L / 310K = 0.01L/K)

0.01L/K = X / 340K

(multiply 340K on both sides, it cancels out on the right)

0.01L/K * 340K = X

(0.01L/K * 340K = 3.29L)

**3.29L = X**

[or]

**3.29L = V2**

Answer:

the relationship between independent and dependent variable

Explanation:

A straight line or linear graph is one of the ways to represent a given data. It shows the relationship between two given set of data; one called the independent variable is plotted on the x-axis (horizontal) while the other called the dependent variable is plotted on the y-axis (vertical).

The straighter the line is, the stronger the relationship between the two variables and vice versa. Hence, the straight line in the graph represents the relationship between independent and dependent variable.

Answer:

one mole of water (6.022 x 10 23 molecules) has a mass of 18.02 g. One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.

Explanation:

• The mole (or mol) represents a certain number of objects.

• SI def.: the amount of a substance that contains the same

number of entities as there are atoms in 12 g of carbon-12.

• Exactly 12 g of carbon-12 contains 6.022 x 10 23 atoms.

• One mole of H 2O molecules

contains 6.022 x 10 23 molecules.

• 1 mole contains 6.022 x 10 23 entities (Avogadro’s number)

• One mole of NaCl contains 6.022 x 10 23 NaCl formula units.

• Use the mole quantity to count formulas by weighing them.

• Mass of a mole of particles = mass of 1 particle x 6.022 x 1023

Mass of 1 H atom: 1.008 amu x 1.661 x10-24 g/amu = 1.674 x10-24 g

Mass of 1 mole of H atoms:

1.674 x10-24g/H atom x 6.022 x1023H atoms = 1.008 g  

• The mass of an atom in amu is numerically the same

as the mass of one mole of atoms of the element in grams.

• One atom of sulfur has a mass of 32.07 amu;

one mole of S atoms has a mass of 32.07 g

Answer:

a. 7.24m

b. 5.15M

c. 53.4mL of the solution would contain this amount of ethanol.

Explanation:

Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.

Molarity, M, are the moles of solute per kg of solvent

To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:

a. Moles ethanol -Molar mass: 46.07g/mol-:

25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol

kg solvent:

100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg

Molality:

0.54265 moles ethanol / 0.075kg = 7.24m

b. Liters solution:

100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L

Molarity:

0.54265 moles ethanol / 0.1053L = 5.15M

c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =

53.4mL of the solution would contain this amount of ethanol

Answer:

A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.

Explanation:

You have to put energy into a molecule to break its chemical bonds. The amount needed is called the bond energy. After all, molecules don't spontaneously break

Answer:

111.95mL of HNO3 are needed to prepare the buffer

Explanation:

We can solve this equation using H-H equation for bases:

pOH = pKb + log [HA+] / [A]

Where pOH is the pOH of the solution

pOH = 14 - pH = 14 - 8.970 = 5.03

pKb is the pKb of NH3 = 4.74

[HA+] could be taken as moles of NH4+

[A] as moles of NH3

The NH3 reacts with nitric acid, HNO3, as follows:

NH3 + HNO3 → NH4+ + NO3-

That means the moles of HNO3 added = X = Moles of NH4+ produced

And moles of NH3 are initial moles NH3 - X

Initial moles of NH3 are:

0.125L * (0.374mol/L) = 0.04675 moles NH3

Replacing in H-H equation:

pOH = pKb + log [HA+] / [A]

5.03 = 4.74 + log [X] / [0.04675-X]

0.29 = log [X] / [0.04675-X]

1.95 =  [X] / [0.04675-X]

0.0912 - 1.95X = X

0.0912 = 2.95X

X = 0.0309 moles

We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:

0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed

In mL:

111.95mL of HNO3 are needed to prepare the buffer

Answer:

c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.

Answer:

Mg> H> Cu

Explanation:

We can see from the question that hydrochloric acid reacted with magnesium as follows;

Mg(s) + 2HCl(aq) ----> MgCl2(aq) + H2(g)

Copper does not react with HCl which means that copper is less reactive than hydrogen hence it can not displace hydrogen from a dilute acid solution.

The order of reactivity of the elements then is ; Mg> H> Cu

Answer:

10.2 mg

Explanation:

Step 1: Calculate the total amount of water she drank

1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.

If she drank 1.4 L of water per day, the total amount of water she drank is:

730 day × 1.4 L/day = 1022 L

Step 2: Calculate the amount of Pb in 1022 L of water

The concentration of Pb is 10 ppb (10 μg/L).

1022 L × 10 μg/L = 10220 μg

Step 3: Convert 10220 μg to milligrams

We will use the conversion factor 1 mg = 1000 μg.

10220 μg × 1 mg/1000 μg = 10.2 mg

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= [tex]\frac{100.0 \ g}{114.33\ g/mole}[/tex]

= [tex]0.8747 \ moles[/tex]

We know that 16 moles of [tex]CO_2[/tex] formation associates with -11018 kJ of heat, then

0.8747 moles [tex]CO_2[/tex] formation associates with,

= [tex]-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat[/tex]

= [tex]-0.0547\times 11018[/tex]

= [tex]-602.69 \ KJ \ heat[/tex]

Answer:

hydrogen

Explanation:

The gas with the least molecular weight effuses the fastest (Graham's Law). Hence, H gas has a higher rate of diffusion compared to N, O, and Cl.

So, Cl is the slowest when it comes to the rate of diffusion, because it has the highest molecular weight.

[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]

Balanced Equation:-

[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]

Answer:

Where the products are H2O and Ba(NO3)2

Explanation:

A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called neutralization reaction.

The balanced reaction is:

Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2

Where the products are H2O and Ba(NO3)2

Answer:

25 g

Explanation:

Step 1: Write the balanced equation

HNO₃ + NaHCO₃ ⇒ NaNO₃ + H₂O + CO₂

Step 2: Calculate the reacting moles of HNO₃

100.0 mL of 3.0 mol/L HNO₃ reacted.

0.1000 L × 3.0 mol/L = 0.30 mol

Step 3: Calculate the reacting moles of NaHCO₃

The molar ratio of HNO₃ to NaHCO₃ is 1:1. The reacting moles of NaHCO₃ are 1/1 × 0.30 mol = 0.30 mol.

Step 4: Calculate the mass corresponding to 0.30 moles of NaHCO₃

The molar mass of NaHCO₃ is 84.01 g/mol.

0.30 mol × 84.01 g/mol = 25 g

Answer:

2

Explanation:

i did this

Answer:

Both Tech A and Tech B.

Explanation:

Catalyst is an element used to start chemical reaction but is not used in the reaction. Catalysts material used in catalytic converter include Rhodium, Palladium and platinum. The pre cat and post cat oxygen sensors helps determine converter efficiency.

Answer:

Explanation:

B

The equilibrium constant for the reaction is K = 0.137

We obtain the equilibrium constant considering the following equilibria and their constants:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

COO(s) + CO(g) → Co(s) + CO₂(g)   K₂ = 490

We write the first reaction in the forward direction because we need H₂(g) in the reactants side:

(1)     COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):

(2)   Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

From the addition of (1) and (2), we obtain:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

+

Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

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

H₂(g) +  CO₂(g) → CO(g) + H₂O(g)

Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.

Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:

K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137

You can learn more about equilibrium constants here:

https://brainly.com/question/15118952

Answer:

26.4g

Explanation:

The balanced chemical equation as stated in this question is given as follows:

Fe2O3 + 3CO → 2Fe + 3CO2

According to this balanced equation, 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).

We need to change 50.26 g of ferric oxide to moles by using the formula;

mole = mass/molar mass

Molar mass of Fe2O3 = 56(2) + 16(3)

= 112 + 48

= 160g/mol

mole = 50.26/160

mole = 0.314mol of Fe2O3

If 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).

Hence, 0.314 mol of Fe2O3 will completely react with (0.314 × 3) mol of CO

0.314 × 3 = 0.94 mol of CO

molar mass of CO = 12 + 16 = 28g/mol

mole = mass/molar mass

mass = mole × M.M

mass = 0.94 × 28

mass = 26.4g of CO

Answer:

λ = 150 nm

Explanation:

For C-O bond rupture:

The required energy to rupture C-O bond = bond energy of C-O bond

= 799 kJ/mol

[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]

[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]

[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]

Recall that the wavelength associated with energy and frequency is expressed as:

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]\lambda = \dfrac{hc}{E}[/tex]

[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]

[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]

λ = 150 nm

Answer:

a. 1.12 L

Explanation:

Step 1: Write the balanced equation for the photosynthesis

6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)

Step 2: Calculate the moles corresponding to 2.20 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

2.20 g × 1 mol/44.01 g = 0.0500 mol

Step 3: Calculate the moles of O₂ produced

The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol

Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP

At STP, 1 mole of O₂ occupies 22.4 L.

0.0500 mol × 22.4 L/1 mol = 1.12 L