What are TWO control variables?

what might cause the goldfish population to drop if the temperature moves out of its normal range? Name three things

Slav-nsk [51]

Answer:

Appearing disoriented, such as swimming upside down.

Leaving food uneaten.

White spots on fins or body.

Discolored gills.

Explanation:

Gold Fish are unable to regulate their body temperature, so they are influenced by the temperature around them. If the water is warm, gold fish metabolism accelerates, feeding and respiration increases, and there is a general increase in movement. If the water is cooler, fish become lethargic and tend to be inactive.

8 0

3 years ago

g How many turns through the citric acid cycle are required to fully oxidize all of the acetly-coA that result from 1 molecule o

faust18 [17]

Answer:

The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.

Explanation:

Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for the reactions of glycolsis is given below:

Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺

Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.

2 Pyruvate ----> 2 AcetylCoA + 2CO₂

Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.

7 0

3 years ago

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

V₂ = 4,00L

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

V₂ = 16,00L

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

V₂ = 12,31L

I hope it helps!

6 0

4 years ago

Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying

PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ( $K_a$ for HF is $6.8\times 10^{-4}$ .)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = $K_a=6.8\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c 0 0

At equilibrium :

(c-x) x x

The expression of disassociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$K_a=\frac{x\times x}{(c-x)}$

$6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}$

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

$[H^+]=x=0.01346 M$

The pH of the solution is ;

$pH=-\log[H^+]=-\log[0.01346 M]=1.87$

The pH of an 0.280 M HF solution is 1.87.

6 0

3 years ago

Which is the best practice recommended in the safety video to mix and acid or a base with a solvent?

Vesna [10]

Answer:

Never pour water into acid but acid into water

Explanation:

If water is poured into extremely concentrated acid/bases, the rate of volatility and exothermic reaction is too rapid and might cause a chemical eruption, leading to acid burns.

Safety precautions hence dictate the reverse is practiced.

I believe this is a clear answer.

4 0

4 years ago