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2 years ago

If u can write like a sentence or two would be amazing:

1 answer:

mihalych1998 [28]2 years ago

7 0

Respiration is the process of oxidation of food materials like glucose, Amino acids, fatty acids to water and carbon dioxide. Respiration does not require any external heat to carry on the process. Combustion is only done by applying external heat to the sugar molecule to burn.

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A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$

Now, find moles of $CH_{3} OH$ are needed using the molarity equation:

$M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol$

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M

5 0

3 years ago

Which of the following reactions have a positive ΔSrxn? Check all that apply.

PolarNik [594]

Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

2A(g) + B(s) → 3C(g)

2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

2A(g) + B(s) → 3C(g)

1) The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

A(g) + B(g) → C(g)

Again, 2 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.

6 0

3 years ago

How does the density of air affect local weather patterns?

JulijaS [17]

When winds move air masses, they carry their weather conditions (heat or cold, dry or moist) from the source region to a new region. When the air mass reaches a new region, it might clash with another air mass that has a different temperature and humidity. This can create a severe storm.

5 0

3 years ago

The process that splits rock through freezing and thawing is called?

maw [93]

It's called Ice Wedging :)

8 0

3 years ago

In the peptide GLVIW, the C-terminal end is ________.

zheka24 [161]

The peptide given above is made up of the following amino acids: glycine [G], leucine [L], valine [V], isoleucine [I] and tryptophan [W]. These amino acids are joined together by amide bond to form peptide. Peptides usually have two terminals, the N terminal and the C terminal. For GLVIW, the C terminal end amino acid is tryptophan, that is the last amino acid on the peptide chain. The N terminal amino acid is glycine, that is, the first amino acid on the peptide chain.

5 0

2 years ago