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2 years ago

5

BRAINLIEST FOR THE FIRST CORRECT ANSWER!! im revising for a test and my teacher gave me some questions to do but i dont understa

nd them. could you pls help me out

1 answer:

il63 [147K]2 years ago

3 0

Answer:

tensile strength, elongation, flexural strength, and impact resistance.

Explanation:

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Name at least four other gases in the atmosphere besides oxygen and nitrogen

nikitadnepr [17]

Carbon dioxide (CO2)
Argon (Ar)
Hydrogen (H)
Helium (He)

8 0

2 years ago

The pKa of the α‑carboxyl group of serine is 2.21 , and the pKa of its α‑amino group is 9.15 . Calculate the average net charge

sleet_krkn [62]

Answer:

Net charge in serine at pH equal to 8.30 is "0"

Explanation:

At pH > $pK_{a}$ , carboxyl group exists as $-CO_{2}^{-}$ (charged)

At pH < $pK_{a}$ , carboxyl group exists as -COOH (neutral)
At pH > $pK_{a}$ , amino group exists as $-NH_{2}$ (neutral)
At pH < $pK_{a}$ , amino group exists as $-NH_{3}^{+}$ (charged)
So, at pH = 8.30, both carboxyl and amino group exists in charged state.
Net charge in serine at pH equal to 8.30 is "0".
Structure of serine at pH equal to 8.30 has been shown below.

8 0

3 years ago

Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re

bearhunter [10]

Answer: 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles$

The balanced chemical equuation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

According to stoichiometry :

4 moles of $Al$ produce == 2 moles of $Al_2O_3$

Thus 0.556 moles of $Al$ will produce= $\frac{2}{4}\times 0.556=0.278moles$ of $Al_2O_3$

Mass of $Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g$

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0

2 years ago

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

<u>Chemical reaction involved</u>: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>

4 0

3 years ago

The isomerization reaction, ch3nc → ch3cn, is first order and the rate constant is equal to 0.46 s-1 at 600 k. what is the conce

Harlamova29_29 [7]

According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3

6 0

3 years ago