Identify the true statements about introns.a- they code for polypeptide proteinsb- they have a branch site located 20 to 50 nucleotides upstream of the 3' splice sitec- they end with the nucleotides AG at the 3' endd- they begin with the nucleotides GU at the 5' ende- they tend to be common in bacterial genes

Answers

Answer 1

Answer:

Answer:

The answer is "Option b, c, and d".

Explanation:

In such a gene, Autosomes are also the sequence for code and transposable elements, not the series of encoding. Through the expression of genes, such fragments of its introns are split through protein complexes throughout the translation process. There has been no kenaf fiber in the genomes of prokaryotic cells.

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Iridium-192 is used in medicine to treat prostate cancer. Iridium-192 has two modes of radioactive decay: 96% of the time it decays by beta emission and 4% of the time it decays by electron capture. What are the daughter nuclides of these two decay processes?
Write the balanced chemical equation for each of these reactions. Include phases.1) When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms.2) However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion
What is the electron pair geometry for a phosphine molecule, PH3? A) tetrahedral B) bent C) linear D) trigonal pyramidal E) none of the above
What is the mole fraction of calcium chloride in 3.35 m CaCl2(aq)? The molar mass of CaCl2 is 111.0 g/mol and the molar mass of water is 18.02 g/mol.
A cube of aluminum measures 3 cm on each side and weighs 81 g. What is its density? (HINT: find the volume of the block first)D=M=V=

Assuming that the distances between the two ions are the same in all cases, which of the following ion pairs has the greatest electrostatic potential energy (i.e., largest in magnitude)? Please explain your answer.a.) Na+ - Cl- b.) Na+ - O-2. c.) Al+3 - O-2. d.) Mg+2-O-2 e.) Na- -Mg+2

Answers

Answer:

Correct option: C

Explanation:

As given in the question that distance between two ions are same in all cases hence r is same for all.

potential energy:

$P.E =(k* q_(1) * q_(2))/(r)$

therefore potential energy depend on the two charge muliplication

so higher the charge multiplication higer will be the potential energy.

magnitude of charge multiplication follow as:

a. 1

b. 2

c. 6

d. 4

e. 2

in option C it is higher

so correct option is C

Identify each element below, and give the symbols of the other elements in its group. a. [Ar] 4s23d104p4
b. [Xe] 6s24f145d2
c. [Ar] 4s23d5.

Answers

Answer:

Explanation:

the electron configuration is defined as the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is

used to describe the orbitals of an atom in its ground state

The valence electrons, electrons in the outermost shell, can be used to know the chemical property

Chemical Name of the Element: Selenium

Chemical Symbol: Se

Group it belong in periodic table:6A

Other Element in the same group:tellurium(Te),,sulfur(S)

atomic number = 34

Selenium is a chemical element that has symbol Se It is a nonmetal which is usually classified as metalloid with properties that are intermediate between the elements above and below in the periodic table.

b)Chemical Name of the Element:Hafnium

Chemical Symbol: Hf

Group it belong in periodic table:4B

Other Element in the same group: Titanium( Ti )Rutherfordium

atomic number: 72

Hafnium is a solid at room temperature.

c)Chemical Name of the Element: Manganese

Chemical Symbol:Mg

Group it belong in periodic table:Mn

Other Element in the same group:Bohrium(Bh) ,Technetium(Tc)

A substance with a high vapor pressure at normal temperatures is often referred to as volatile. As the temperature of a liquid increases, the kinetic energy of its molecules also increases and as the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure.

Answers

As the temperature of a liquid or stable will increase its vapor strain additionally will increase. Conversely, vapor strain decreases because the temperature decreases.

The better the vapor strain of a substance, the extra the awareness of the compound withinside the gaseous section and the extra the quantity of vaporization

. Liquids range substantially of their vapor pressures. substance with a excessive vapor strain at everyday temperatures is regularly called volatile. The strain exhibited through vapor gift above a liquid floor is referred to as vapor strain. As the temperature of a liquid will increase, the kinetic strength of its molecules additionally will increase.

Answers

Answer:

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 12CO2 + 6H2O

The coefficients are 2, 15, 12, 6

2b. 126.25g of CO2

Explanation:

1a. Step 1:

Equation for the reaction. This is given below:

NO + O2 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → NO2

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:

2NO + O2 → 2NO2

The equation is balanced.

The coefficients are 2, 1, 2

1b. Step 1:

Determination of the limiting reactant. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO required 1 mole of O2.

Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.

From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.

1b. Step 2:

Determination of the maximum amount of NO2 produced. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO produced 2 moles of NO2.

Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.

1b. Step 3:

Conversion of 16.42 moles of NO2 to grams. This is illustrated below:

Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol

Mole of NO2 = 16.42 moles

Mass of NO2 =?

Mass = number of mole x molar Mass

Mass of NO2 = 16.42 x 46

Mass of NO2 = 755.32g

Therefore, the maximum amount of NO2 produced is 755.32g

2a. Step 1:

The equation for the reaction.

C6H6 + O2 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:

C6H6 + O2 → 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 → 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 12CO2 + 6H2O

Now, the equation is balanced.

The coefficients are 2, 15, 12, 6

2b. Step 1:

Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol

Mass of C6H6 from the balanced equation = 2 x 78 = 156g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 15 x 32 = 480g

2b. Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

156g of C6H6 required 480g of O2.

Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.

From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.

2b. Step 3:

Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb

2C6H6 + 15O2 → 12CO2 + 6H2O

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

Mass of CO2 from the balanced equation = 12 x 44 = 528g

2b. Step 4:

Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:

From the balanced equation above,

156g of C6H6 produced 528g of CO2.

Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2

300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

Answers

Answer: The new concentration of the solution is 0.143 M.

Explanation:

Given: $V_(1)$ = 300.0 mL, $M_(1)$ = 0.335 M

$V_(2)$ = 700.0 mL, $M_(2)$ = ?

Formula used is as follows.

$M_(1)V_(1) = M_(2)V_(2)$

Substitute values into the above formula as follows.

$M_(1)V_(1) = M_(2)V_(2)\n0.335 M * 300.0 mL = M_(2) * 700.0 mL\nM_(2) = 0.143 M$

Thus, we can conclude that the new concentration of the solution is 0.143 M.

Final answer:

To find the new concentration of the solution, you can use the formula C1V1 = C2V2. Plugging in the given values, the new concentration of the solution is 0.144 M.

Explanation:

To find the new concentration of the solution, we can use the formula:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Plugging in the given values, we get:

(0.335 M)(300.0 mL) = C2(700.0 mL)

Solving for C2, we find the new concentration of the solution to be 0.144 M.

Learn more about concentration of a solution here:

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A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is 4.70 4.70 , what is the pH of the buffer?