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Natalka [10]
3 years ago
9

Which of the following systems in not directly involved in the body's regulation of water?

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
7 0

The muscular system is the only system that does not directly involved in the body's regulation of water. So lettered B is the answer.

Have a good day!

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Which statement describes the masses of the particles that make up an atom?
Novay_Z [31]
The mass of the atom is equal to the sum of the number of protons and the number of neutrons. In a neutral atom, the number of protons is equal to the number of electrons.  The atomic number meanwhile of an atom is equal to the number of protons of the atom.
6 0
3 years ago
Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and
alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

                                = 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

7 0
2 years ago
How many atp molecules are produced during the citric acid cycle?
Mrrafil [7]
"Only 2 molecules" of ATP <span>produced during the citric acid cycle

Hope this helps!

</span>
7 0
3 years ago
Read 2 more answers
Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
melisa1 [442]

Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

7 0
3 years ago
Calculate the final Celsius temperature when 634 L at 21 °C is compressed to 307 L.
Illusion [34]

Answer:

- 130.64°C.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.

V₂ = 307.0 L, T₂ = ??? K.

<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>

<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>

5 0
3 years ago
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