What’s my resting energy????

What is the answer please and why is this answer

Snowcat [4.5K]

Answer:

it is B because the other answers logically dont fit in

5 0

3 years ago

The reduction in the freezing point of a solution is inversely proportional to a molal concentration

maks197457 [2]

The statement above about "The reduction in the freezing point of a solution is inversely proportional to a molal concentration" is false. It must be directly proportional

7 0

4 years ago

PLEASE HELP WILL MARK AS BRAINLEST

zlopas [31]

Answer:

The deer population will grow

This is because the Moutain Lions were eating the deer and now nothing is eating them.

8 0

3 years ago

In an experiment, a student made a compund of sulfur and copper, and now she needs to find if the compound is copper (I) sulfide

finlep [7]

Answer: The formula for the given compound is $Cu_2S$

Explanation:

An ionic compound is defined as the compound which is formed when electron gets transferred from one atom to another atom. These are usually formed when a metal reacts with a non-metal.

The chemical formula of copper (I) sulfide is $Cu_2S$ because this is formed by the combination of $Cu^+$ ions and $S^{2-}$ ions

The chemical formula of copper (II) sulfide is $CuS$ because this is formed by the combination of $Cu^{2+}$ ions and $S^{2-}$ ions

To calculate the percent composition of element in a compound, we use the equation:

$\text{Percent composition of element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$ ...........(1)

For copper (I) sulfide:

Mass of copper = (2 × 63.546) = 127.092 g/mol

Mass of copper (I) sulfide = 159.16 g/mol

Putting values in equation 1, we get:

$\text{Percent composition of copper in }Cu_2S=\frac{127.092g}{159.16g}=79.85\%$

For copper (II) sulfide:

Mass of copper = 63.546 g/mol

Mass of copper (II) sulfide = 95.611 g/mol

Putting values in equation 1, we get:

$\text{Percent composition of copper in CuS}=\frac{63.546g}{95.611g}=66.46\%$

Given compound:

Mass of copper in given compound = 0.8383 grams

Mass of given compound = 1.0440 grams

Putting values in equation 1, we get:

$\text{Percent composition of copper in given compound}=\frac{0.8383g}{1.044g}=80.29\%$

The percent composition of copper is close to the composition in copper (I) sulfide.

Hence, the formula for the given compound is $Cu_2S$

8 0

3 years ago

Your company has developed an organic molecule with commercial potential and you know how to produce it in the lab. You want to

Lena [83]

Answer:

(B) continually remove products

(C) increase the concentration of reactants

Explanation:

You can drive the production rate of an equilibrium reaction by handling several factors: temperature, reactant concentrations, and product concentrations are the main of those factors.

The thermodynamic variable that tells whether a chemical reaction is spontaneous is the free energy, ΔG:

ΔG < 0 represents a driving force in the forward direction,

ΔG > 0 represents a driving force in the reverse direction,

ΔG = 0 represents that the system is at equilibrium.

ΔG is related with ΔG° per the expression:

ΔG = ΔG° + RT ln Q

Where Q represents the ratio between the molar concentration of the products (each raised to its stoichiometric coefficient) and the molar concentrations of the reactants (each raised to its stoichiometric coefficient).

Since, you want to increase your production, means you want to favor the forward reaction and that means that you want to make ΔG more negative.

So, you want to make the term RT lnQ more netative.

Logarithm function of a rational expression gets more negative when the numerator decreases or the denominator increases.

So, you want either reduce the amount of products or increase the amount of reactants, which is given by the options B and C:

(B) continually remove products

(C) increase the concentration of reactants

Enzymes are a kind of catalyst. In an equilibrium reaction, a catalys speeds up both the forward and reverse reaction rates equally, so the equilibrium concentrations will not change. So, adding an enzyme (choice A) would help if you, continually remove products (B) or increase the concentration of reactants (C).

Adding products to get the reaction primed (D) will not help because that just would drive to the consumption of part of the products to obtain some reactants until reaching the equilibrium.

8 0

3 years ago