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3 years ago

In the polypeptide Phe-Tyr-Glu-Asp-Ser-Ile-Leu-Ser what is the N-terminal amino acid?

Chemistry

1 answer:

avanturin [10]3 years ago

7 0

Answer:

N-terminal Phe, C-terminal Ser

Explanation:

Amino acids connect like

NH2 -CH(R1) -CO -NH-CH(R2)-CO-.....-NH-CH(Rn)COOH

So, 1st amino acid is N -terminal , and it is Phe.

Last amino acid is C- terminal, and it is Ser.

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3 years ago

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A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L

NeTakaya

Answer:

$\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}$

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

$P_1V_1=P_2V_2$

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

$25.0 \ L * 2.05 \ atm = P_2V_2$

The volume is decreased to 14.5 liters, but the pressure is unknown.

$25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L$

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}$

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2$

The units of liters cancel.

$\frac {25.0 * 2.05 \ atm }{14.5 }=P_2$

$\frac {50.25\ atm }{14.5 }=P_2$

$3.53448276 \ atm = P_2$

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

$3.53 \ atm \approx P_2$

The new pressure is approximately <u>3.53 atmospheres.</u>

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2 years ago

Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2

Flauer [41]

Answer:

$\boxed{28.81}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 58.12 44.01

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g: 9.511

1. Moles of C₄H₁₀

$\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}$

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

$\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}$

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$\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}$

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2 years ago

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Answer:

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2 years ago

In order for a nonmetal to obey the octet rule, it____.

lubasha [3.4K]

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3 years ago

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