A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of___________ bases along its backbone. a. 146 b. 292 c. 438 d. 20 e. 3

Answers

Answer 1

Answer:

Since the relationship between amino acid and codon bases would be the values of 3 nitrogenous bases per 1 amino acid.

knowing this relationship what you would do is simply multiply 146 x 3 to find the number of codon bases which would be C. 438.

Answer 2

A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.

What is messenger RNA?

mRNA or messenger RNA is a single stranded RNA molecule. It is complementary to the DNA and carries genetic information present in the DNA. It is translated to form proteins. The genetic codes (triplet) present on mRNA get translated to amino acids, giving rise to the functional product of a gene.

So mRNA really is a form of nucleic acid, which helps the human genome which is coded in DNA to be read by the cellular machinery. mRNA is actually the translated form of DNA that the machinery can recognize and use to assemble amino acids into proteins.

Each strand has 3 bases, so 146 × 3 = 438 bases

Therefore, A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.

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Related Questions

Calculate a pressure at a point that is 100 m below the surface of sea water of density 1150 kgm?

Answers

Answer:

1229.08975 kPa

Explanation:

Given that:

The depth of the water = 100 m below the surface of the water.

The density of the water = 1150 kgm

The mass of the water = depth of the water × density of the water

The mass of the water = 100 m × 1150 kgm

The mass of the water = 115000 kgf/m²

The mass of the water is the pressure of the water at a depth of 100 m below the surface of the water.

Since 1 kgf/m² = 0.00980665 kPa

Then  115000 kgm² = (115000  ×  0.00980665) kPa

=  1127.76475 kPa

At standard temperature and pressure , the atmospheric pressure = 101.325 kPa.

Therefore, the pressure below the surface of sea water  = 1127.76475 kPa +   101.325 kPa

= 1229.08975 kPa

3. Strontium-90 is produced during the nuclear fission of uranium-235 and is part of nuclear fallout created by weapons testing. If the half-life of Sr-90 is 28 days, how long will it take for grass contaminated with Sr-90 to be safe (<2 percent of the starting radioactivity) for cattle to eat?
A. 158 days
B. 28 days
C. 1 year
D. 158 years​

Answers

Answer:158 days (D)

Explanation:

It will take 158 days for grass contaminated with Sr-90 to be safe for cattle to eat. Therefore, option (A) is correct.

What is the half-life period?

The half-life of a radioactive material is defined as the time that is needed to reduce the initial quantity of a radioactive element to half after disintegration.

The half-life of a radioactive element can be described as the characteristic of the element and does not influence by the initial amount of the radioactive substance.

Given, the half-life of the Strontium-90 = 28 days

The rate constant of the decay can be determined as:

[tex]t_{\frac{1}{2} } =\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{t_{\frac{1}{2} } }[/tex]

k = 0.693/28

k = 0.025 day⁻¹

The concentration of Strontium-90 reduced to less than 2% is safe. Therefore final concentration [A] = 2 % = 0.02

[tex]t = \frac{2.303}{k} log \frac{[A_o]}{[A]}[/tex]

[tex]t = \frac{2.303}{0.02475} log \frac{1}{[0.02]}[/tex]

t = 158 days

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Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?

Answers

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

Where the concentrations of the ions are the concentrations in equilibrium

For actual concentrations of a solution, you can define Q, reaction quotient, as:

Q = [Ba²⁺] [CO₃²⁻]

If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

100.0mL is the volume of the mixture of the solutions

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

A precipitate will form, BaCO₃

Which element always exists in monatomic form? silicon, a metalloid oxygen, a nonmetal calcium, a metal neon, a noble gas

Answers

Answer:

[tex]\huge\boxed{A \ noble \ gas}[/tex]

Explanation:

A noble gas always exists in a mono-atomic form. This is because the valency of noble gas is zero and thus they don't tend to lose, gain or share electrons. They are inert gases and do not react.

Answer:

a noble gas

Explanation:

the the noble gases exist as mono atomic elements that is helium, neon ,argon, Krypton,etc.

mono atomic elements exist as stable single atoms

volume of percentage =
formula??​

Answers

Answer:

Volume percent is defined as: v/v % = [(volume of solute)/(volume of solution)] x 100%

Explanation:

What are periodic trends if ionic radii

Answers

Answer:

Explan ionization energy, atomic radius, and electron affinityation:

This question most likely has answer choices. The possible answer choices are as followed:

Ionic radii tend to increase down a group.Ionic radii tend to decrease across a period.Anionic radii tend to increase across a period.Cationic radii tend to decrease across a period.Ionic radii increase when switching from cations to anions in a period.

The answers are Ionic radii tend to increase down a group, Cationic radii tend to decrease across a period, and Ionic radii increase when switching from cations to anions in a period (1st, 4th, and 5th options).

Predict the most likely bond type for the following.

a. Cu (Copper)
b. KCl (Potassium Chloride)
c. Si (Silicon)
d. CdTe (Cadmium Telluride)
e. ZnTe (Zinc Telluride)

Answers

Answer:

The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.

Explanation:

a. Cu (Copper)- ionic bonding

b. KCl (Potassium Chloride) - ionic bonding

c. Si (Silicon) - covalent bonding

d. CdTe (Cadmium Telluride) - polar covalent bonding

e. ZnTe (Zinc Telluride)- polar covalent bonding

Which of the following is not the same as 1,400 mL? a. 1.4 cm³ b 1.4 L c. 1,400 cm³ d. 140 cL

Answers

answer should be 1.4 cm³

1 L = 10 and so

dL = 100 and then

cL = 1,000

mL = 0.001 m³

1 m³ = 1,000

dm³ = 1,000,000

cm³ = 1,000,000,000

mm³ = 1,000 L

So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL

1,400 mL = 1,400 cm³ = 1.4 L = 140 cL

Answer:

1.4 cm^3

Explanation:

The ΔHvap of nitrous oxide is 16.53 kJ · mol−1 and its ΔSvap is 89.51 J · mol−1 · K−1. What it the boiling point of nitrous oxide?

Answers

Answer:

[tex]T_b=-88.48\°C[/tex]

Explanation:

Hello,

In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:

[tex]\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}[/tex]

Solving for the boiling point of nitrous oxide, we obtain:

[tex]T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K[/tex]

Which in degree Celsius is also:

[tex]Tb=184.67-273.15\\\\T_b=-88.48\°C[/tex]

Best regards.

Calculate the pH of a buffer that is 0.13 M in lactic acid and 0.10 M in sodium lactate. Express your answer using two decimal places.

Answers

Answer:

pH of the buffer is 3.75

Explanation:

It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid

In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:

pH = 3.86 + log [Lactate] / [Lactic acid]

Replacing with the concentrations of the problem:

pH = 3.86 + log [0.10M] / [0.13M]

pH = 3.75

pH of the buffer is 3.75

If the rate of formation (also called rate of production) of compound C is 2M/s in the reaction A --->2C, what is the rate of consumption of A

Answers

Answer:

[tex]r_A=-1\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:

[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]

In such a way, solving the rate of consumption of A, we obtain:

[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]

Clearly, such rate is negative which account for consumption process.

Regards.

Which of the following elements can't have an expanded octet? answers A. oxygen B. phosphorous C. chlorine d. sulfer

Answers

answer is oxygen .

oxygen is an exception in octet rule

Among the following given elements,oxygen is an element which cannot have an expanded octet.

What is an expanded octet?

Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds  are not less common  and are of equal stability as the compounds which obey octet rule.

Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.

Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does  not have the 2d -orbitals   due to which it  does not fulfill the criteria of an element to have an expanded octet.

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Enter your answer in the provided box.


The equilibrium constant KP for the reaction


CO(g) + Cl2(g) ⇌ COCl2(g)


is 5.62 × 1035 at 25°C. Calculate ΔG

o

f

for COCl2 at 25°C.

Answers

Answer:

The correct answer is -341.2 kJ per mole.

Explanation:

The reaction given is:  

CO (g) + Cl₂ (g) ⇔ COCl₂ (g)

Kp = 5.62 × 10³⁵

T = 25 °C or 298 K

The formula for calculating ΔG is,  

ΔG° = -RTlnKp

ΔG° = -8.314 × 298 ln (5.62 × 10^35)

ΔG° = -203.9 kJ/mol

ΔG° = ∑nΔG°f (products) -∑nΔG°f (reactants)

ΔG° = ΔG°f (COCl₂ (g)) - [ΔG°f (CO(g)) + ΔG°f (Cl₂(g))]

ΔG°f (COCl₂ (g)) = ΔG° + [ΔG°f (CO (g)) + ΔG°f (Cl₂(g))]

ΔG°f (COCl₂ (g)) = -203.9 + (-137.28 + 0.00)

ΔG°f (COCl₂ (g)) = -341.2 kJ/mol

The standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol

The given equation for the chemical reaction is

CO(g) + Cl2(g) ⇌ COCl2(g)

At the temperature of 25°C = (273 + 25) K, the equilibrium constant [tex]\mathbf{K_p = 5.62\times 10^{35}}[/tex]

Consider the expression for the relationship between [tex]\mathbf{\Delta G^o}[/tex] and [tex]\mathbf{K_p }[/tex] for the equilibrium reaction can be expressed as:

[tex]\mathbf{\Delta G^o = - RT In K_p}[/tex]

where;

gas constant (R) = 8.314 × 10⁻³ kJ/K.mol

[tex]\mathbf{\Delta G^o = - (8.314 \times 10^{-3}\ kJ/K.mol \times 298 \ K) \times In (5.62 \times 10^{35} )}[/tex]

[tex]\mathbf{\Delta G^o = -2.477572\ K \times 82.31680992}[/tex]

[tex]\mathbf{\Delta G^o = 203.95 \ kJ}[/tex]

Thus, the standard free energy for the reaction is 203.95 kJ/mol

For a given reaction, the standard Gibbs free energy can be calculated by using the formula:

[tex]\mathbf{\Delta G^o_{rxn} = \sum n \Delta G^o_f (products) - \sum m \Delta G^o_f (reactants) }[/tex]

[tex]\mathbf{\Delta G^o_{rxn} =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(\Delta G^o_{f} (CO)_{(g)} + \Delta G^o_{f} (Cl)_{2(g)} ) \Big ) \Big ] }[/tex]

replacing the values of and solving for COCl2 at standard free energy of formation of substances, we have:

[tex]\mathbf{-203.95 \ kJ/mol =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(-137.3 kJ/mol + 0 \ kJ/mol\Big ) \Big ] }[/tex]

Collecting like terms, we have:

[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -203.95 \ kJ/mol -137.3 kJ/mol }[/tex]

[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -341.25 \ kJ/mol }[/tex]

Therefore, we can conclude that the standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol

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3,3-dibromo-4-methylhex-1-yne​

Answers

Explanation:

see the attachment. hope it will help you...

How to do q solution, qrxn, moles of Mg , and delta Hrxn?

Answers

Answer:

14, 508J/K

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

m = mass of substance (g) = 0.1184g

1 mole of Mg - 24g

n moles - 0.1184g

n = 0.0049 moles.

Also, q = m × c × ΔT

Heat Capacity, C of MgCl2 = 71.09 J/(mol K)

∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)

= 14, 508 J/K/kg

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg

= 1,7117.7472 J °C-1 g-1

∴ ΔHrxn = q/n

=1,7117.7472  ÷ 0.1184

= 14, 508J/K

Why do prices increase when demand for a product is high? Companies know they can make more money by selling fewer products at higher prices. Companies know that people will be willing to spend more to get an in-demand product. Companies take advantage of the demand to make people spend more money on excess products. Companies know they can stop production and still make money on sales.

Answers

Answer:

Companies know that people will be willing to spend more to get an in-demand product.

Explanation:

When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.

Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.

Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.

Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.

Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.

Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.

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Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light? Sort them accordingly. ltems (6 items) (Drag and drop into the appropriate area below)
a. Gamma rays
b. Infrared radiation
c. Ultraviolet liht
d. X-rays
e. Microwaves
f. Radio waves

Answers

Answer:

Higher frequency than visible light -  Ultraviolet light, X-rays, and Gamma rays

Lower frequency than visible light - Infrared radiation, microwaves, and Radio waves

Explanation:

The frequencies of electromagnetic radiations vary according to their wavelengths. The relationship between the frequency and wavelength of the waves is expressed such that:

      λ = c/f, where λ = wavelength, c = speed of light, and f = frequency.

Thus, there is an inverse relationship between the wavelength and the frequency of electromagnetic waves.

The order of the electromagnetic waves based on their frequency from the lowest to the highest is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma-rays

Hence, electromagnetic waves with higher frequencies than visible light include ultraviolet light, X-rays, and Gamma rays while those with lower frequencies include Infrared radiation, microwaves, and Radio waves.

Answer:

need points

Explanation:


How has the work of chemists affected the environment over the years?

Answers

Answer:

Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.

Answer:

Chemists have both hurt the environment and helped the environment by their actions.

Explanation:

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Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.

Answers

Answer:

whats the ph  ofpoh=9.78

Explanation:

Which one of the following compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide?
a. butanal
b. 2-methylbutanal
c. 3-methylpentanal
d. 2, 2-dimethylbutanal

Answers

It has to be b. 2-methylbutanal

The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.

What is aldol reaction?

The Aldol Reaction occurs when the enolate of an aldehyde or ketone combines with the carbonyl of another molecule at the aplha-carbon under basic or acidic circumstances to produce beta-hydroxy aldehyde or ketone.

For the formation of enolate ion, compound should contain alpha hydrogen in it and among the given compound only 2, 2-dimethylbutanal will not have alpha hydrogens.

Butanal, 2-methylbutanal and 3-methylpentanal will have aplha hydrogens in it so that it takes part in the aldol reaction.

Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.

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A student puts a glass of water in the freezer. Later, he notices ice forming on the surface of the water. Which property of water best explains why ice forms on its surface? A. It is made of polar molecules. B. It has low surface tension. C. It has weak adhesion. D. It is densest as a solid.

Answers

ℯ ℴ ℴ ℴℯℯ

it has a weak adhesion

Which of the following goes through the largest volumetric change? Question 4 options: A) Water when it's heated from 1oC to 99oC B) Water when it freezes into ice C) Ice when it melts into water D) Water when it boils into steam

Answers

Answer:

Water when it freezes into ice

Explanation:

Most liquids expand when heated and contract when cooled, water behaves in an anomalous fashion. Water rather expands when cooled and contracts when heated.

Water usually contracts on cooling from any temperature until 4°C, after 4°C, the water begins to expand rapidly. Hence water has its least volume at 4°C and increases rapidly afterwards.

Thus the largest volume change for water occurs during freezing since it expands when cooled.

It would require ? Liters of water to dissolve 36 grams of the substance.

Answers

The correct answer is 3 liters

Explanation:

If a substance has a solubility of [tex]12 \frac{grams}{liter}[/tex], this means in 1 liter, the grams that can be dissolved are 12 grams. Now, considering Justin and Ellie need to dissolve 36 grams to calculate the number of liters just divide the total of grams into 12 as each liter dissolves only 12 grams. The process is shown below:

36 grams (the amount that will be dissolved) ÷ 12 (grames dissolved per liter) = 3 liters (liters to dissolved 36 grams)

Answer:

It would be 3 liters

Explanation:

For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.

Answers

Answer:

The ΔG° is 29 kJ and the reaction is favored towards reactant.

Explanation:

Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,  

ΔG° = ΔH°rxn - TΔS°rxn

= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K

= 41.2 kJ - 12.2 kJ

= 29 kJ

As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.  

Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ

Answers

Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.

Answers

Answer:

There are no forces holding the gas particles together.

Explanation:

A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbonate that has twice the mass of the first sample

Answers

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . What mass of ammonium phosphate is produced by the reaction of of ammonia?

Answers

Answer:

The mass of ammonium phosphate produced is 14.3g

Explanation:

Full question contains: "What mass of ammonium phosphate is produced by the reaction of 4.9g of ammonia"

Ammonium phosphate ((NH₄)₃PO₄) can be produced by the reaction of phosphoric acid (H₃PO₄) with ammonia (NH₃) as follows:

H₃PO₄ + 3NH₃ → (NH₄)₃PO₄

Where 1 mole of phosphoric acid reacts with 3 moles of ammonia producing 1 mole of ammonium phosphate.

To know how many grams of ammonium phosphate we need to find moles of ammonia that react, and, with the chemical equation we can find moles of ammonium phosphate and its mass as follows:

Moles ammonia (Molar mass: 17.031g/mol):

4.9g × (1mol / 17.031g) = 0.288 moles of ammonia you have in 4.9g

Moles of ammonium phosphate (149.09g/mol) and its mass:

As 0.288 moles of NH₃ are reacting and 3 moles of ammonia produce 1 mole of ammonium phosphate, moles produced are:

Moles (NH₄)₃PO₄:

0.288 moles NH₃ ₓ (1 mol (NH₄)₃PO₄  / 3 mol NH₃) = 0.0959 moles (NH₄)₃PO₄

These moles are, in grams:

0.0959 moles (NH₄)₃PO₄ ₓ (149.09g / mol) = 14.3g ammonium phosphate.

The mass of ammonium phosphate produced is 14.3g

Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond.

Answers

Answer:

See explanation

Explanation:

The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.

The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.

Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.

The structure that should be drawn is shown below.

The reaction of chlorine:

It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.

Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.

Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction

Answers

Answer:

d. Oxidation and reduction

Explanation:

For this question we have to remember the definition of each type of reaction:

-) Hydrogenation

In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.

-) Alkylation

In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.

-) Hydrolysis

In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.

-) Halogenation

In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.

-) Ammoniation

In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.

-) Oxidation and reduction

In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:

[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction

[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation

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