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2 years ago

What is The relationship between a carnivore and an herbivore can be stated as

1 answer:

3 0

Predator and prey because herbivores are animals that only eat plants implying they wouldnt eat the carnivore as the carnivore would eat them because carnivores only eat animals such as the herbivore

not sure if that made sense but its B

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Determine whether the disruption of the bonds or attractions occurs during protein hydrolysis or protein denaturation.

Aleonysh [2.5K]

Answer:

The disruption of the bonds or attractions occurs during protein hydrolysis which results in the loss for the primacy structure. The peptide bonds is the bond affected in this scenario.

The disruption of the bonds however only exist in the process of denaturation and this results in a change in the confirmation which could be secondary, tertiary, and quaternary structural related. And example of the bonds affected include salt bridges, disulfide bridges, hydrogen bonds etc.

5 0

3 years ago

Someone Please HELP Me

scoray [572]

Use a calculator to add those thank you ur welcome

7 0

3 years ago

1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O

elena55 [62]

Answer:

The metal is most likely calcium which has a molar mass of 40.078 g

<em>Note: The question is incomplete. The complete question is given below:</em>

1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O2– ions. The volume of one mole of gas at this temperature and pressure is 24.0dm3 .

What could be the identity of the metal?

<em>A calcium B magnesium C potassium D sodium</em>

Explanation:

Number of moles of the oxygen gas reacted is first determined:

Volume of oxygen gas = 300 cm³ = 0.30 dm³

1 mole of a gas occupies 24.0 dm³ at this temperature and pressure

Number of moles of oxygen gas that will occupy 0.30 dm³ = 0.3/24 = 0.125 moles

Let M be represent the metallic element. The equation of the reaction will be:

2M + O₂ ----> 2MO

From the equation of reaction;

2 moles of the metal reacts with 1 mole of oxygen gas

2 * 0.125 moles of the metal will react with 300 cm³ of oxygen = 0.025 moles

Since the mass of the metal that reacted with oxygen is 1.0 g, therefore, 1.0 g of the metal = 0.025 moles

mass of 1 mole of the metal = 1 /0.025 = 40.0 g

Therefore, the metal is most likely calcium which has a molar mass of 40.078 g

5 0

2 years ago

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .