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1 year ago

Science please help me with this

Chemistry

2 answers:

nataly862011 [7]1 year ago

7 0

Answer:

a lack of genetic variety

first one is the correct one

kvv77 [185]1 year ago

6 0

A lack of genetic variety

The first one

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Silvers symbol on the periodic table of the elements is Ag. what else is true of silver?

pishuonlain [190]

The answer should be D hope this helps

4 0

2 years ago

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.

3 0

2 years ago

Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b.

Ilia_Sergeevich [38]

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) Trypsin

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) Chymotrypsin

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.

3 0

2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

1 year ago

When two atoms form a covalent bond, they share electrons from all of their orbitals. All of their orbitals, in turn, combine to

erma4kov [3.2K]

Answer:

It's false.

Explanation:

Molecular orbital theory states that the number of molecular orbitals is equal to the number of atomic orbitals that overlap. The lowest energy molecular orbital is formed when two atomic orbitals that are in phase overlap, forming a bonding molecular orbital. However, another molecular orbital is also formed, called an anti-binding orbital.

So if an "n" quantity of atomic orbitals is combined, an "n" quantity of molecular orbitals is formed.

Have a nice day!

8 0

2 years ago