8
What happens to solid waste in the circulatory system?
A it's expelled through the lungs
B
It pumps into the coronary circulation
C
It's dropped off in the kidneys
D
It's deposited in the aorta

Answers

Answer 1

Answer:c it’s dropped off in the kidneys

Explanation:

I took the quiz

Answer 2

The solid wastes are from the circulatory system is expelled to the kidney where, the nitrogenous wastes like urea and uric acid is excreted as urine from the body.

What is circulatory system ?

A circulatory system is an organ system, where the blood is purified and oxygenation of blood takes place. Through circulatory system, the blood reaches throughout the body pumped from the heart through veins.

The organs included in circulatory system are lungs, heart, aorta, veins, blood vessels etc. There are various kinds of blood vessels each having specific functions.

There is a network of blood vessels including arteries and large veins, capillaries that join the venules and other veins. All the nutrients and ions are circulated throughout the body through blood and solid wastes are then expelled to kidney.

Kidney function as a sieve to clean the good fluid from waste products. Uric acid and urea along with water excreated as urine then. Thus, option C is correct.

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Related Questions

PLEASE HELP!! ORGANIC CHEMISTRY

A sample of a diatonic gas is loaded into an evacuated bottle at STP. The 0.25 L bottle contains 1.76 grams of the unidentified gas. Calculate the molar mass of the gas. What is the identity of the diatomic gas?

Answers

Answer:

(a) 157.7 g

(b) 7.04 g/dm³

Explanation:

(a) From the question,

According to Avogadro's Law,

1 mole of every gas at STP occupies a volume of 22.4 dm³

But mass of 1 mole of the diatomic gas  = molar mass of the gas.

This Implies that,

The molar mass of the gas at STP occupies a volume of 22.4 dm³

From the question,

If,

0.25 L bottle contain 1.76 g of the gas,

Therefore,

Molar mass of the gas = (1.76×22.4)/0.25

Molar mass of the gas = 157.7 g.

(b) Density of the gas = mass/volume

D = m/v

Given: m = 1.76 g, v = 0.25 L = 0.25 dm³

Therefore,

D = 1.76/0,25

D = 7.04 g/dm³

The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts

Answers

Answer:

E: the point where the acid and base have been added in proper stoichiometric amounts

Explanation:

Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.

The Equivalence point occurs before the endpoint.

Thus, option E is correct.

In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?

Answers

I think you forgot to add a picture?

Answer:

Refraction

Explanation:

Inquiry Extension Consider a reaction that occurs between solid potassium and chlorine gas. If you start with an initial mass of 15.20 g K, and an initial mass of 2.830 g Cl2, calculate which reactant is limiting. Explain how to determine how much more of the limiting reactant would be needed to completely consume the excess reactant. Verify your explanation with an example

Answers

The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].

To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.

The balanced chemical equation for the reaction between solid potassium and chlorine gas is:

2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)

From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.

First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:

moles of K = 15.20 g / 39.10 g/mol = 0.388 mol

moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol

Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:

Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl

Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl

We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.

To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.

We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:

moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])

                                = 0.080 mol K

This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:

mass of K needed = 0.080 mol K x 39.10 g/mol

                              = 3.13 g K

Therefore, The 3.13 g of K would be needed to completely react with the remaining.

Example verification:

Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?

Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]

Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol

Moles of additional [tex]Cl_2[/tex] = 0.0070 mol

The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:

0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl

Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:

5.95 g + 1.04 g = 6.99 g

The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].

Learn more about Solid Potassium at

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Cell membranes are selectively permeable. This means that A. only water can move freely across the cell membrane. B. any substance can move across the cell membrane, but chemical energy will always be required. C. some substances can move freely across the cell membrane, while others must be transported. D. no substances can move freely across the cell membrane.

Answers

Answer:

C. some substances can move freely across the cell membrane, while others must be transported.

Explanation:

explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity

Answers

The answer is… Ions are highly charged species that can readily conduct an electrical current. But hydrogen chloride the gas does not exist as ions since there is no solvent medium for them to dissolve into. Hope this helps!

Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.

Answers

"0.0340" mol of CH₄ are in sealed flask.

Methane (CH₄)

Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.

According to the question,

Volume, V = 800 mL or, 0.800 L

Temperature, T = 22°C or, 295

Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm

As we know the relation,

The gram of moles will be will be:

→ n = [tex]\frac{PV}{RT}[/tex]

By substituting the values, we get

     = [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]

     = [tex]\frac{0.824}{242.077}[/tex]

     = 0.0340

Thus the response above is correct.

Find out more information about methane here:

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A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution

Answers

Answer:

Following are the solution to the given question:

Explanation:

Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.

What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe reacts with 2.1 g of O2?

4 Fe + 3O2 —> 2Fe2O3

Answers

Answer:

Fe is limiting reactant and 3.00g of Fe2O3 will be produced

Explanation:

To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:

Moles Fe -Molar mass: 55.845g/mol-

2.1g * (1mol / 55.845g) = 0.0376 moles

Moles O2 -Molar mass: 32g/mol-

2.1g * (1mol / 32g) = 0.0656 moles

For a complete reaction of 0.0656 moles of O2 are needed:

0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe

As there are just 0.0376 moles,

Fe is limiting reactant

The mass of Fe2O3 is:

Moles:

0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3

Mass:

0.0188 moles Fe2O3 * (159.69g / mol) =

3.00g of Fe2O3 will be produced

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