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Ronch [10]
3 years ago
12

Provide a definition of a closed system.

Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
8 0

Answer:

No idea

Explanation:

I don't know the answer. You're welcome!

tensa zangetsu [6.8K]3 years ago
5 0

Answer:

A closed system is a physical system that does not allow transfer of matter in or out of the system, though, in different contexts, such as physics, chemistry or engineering, the transfer of energy is or is not allowed.

Explanation:

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A compound has the molecular formula C3H7OH. Which class of organic compounds does it belong to?
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8 0
4 years ago
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Convert 2.39 x 1022 molecules of water into moles of water
torisob [31]

Answer:

0.0397\ \text{mol}

Explanation:

Number of molecules of water = 2.39\times 10^{22}

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

Number of moles is given by

\dfrac{2.39\times 10^{22}}{6.022\times 10^{23}}

=0.0397\ \text{mol}

The number of moles of water is 0.0397\ \text{mol}.

6 0
3 years ago
Which mechanisms is most significant in returning the ion concentrations to the resting state?
IceJOKER [234]

Answer:

Active transport by the Na+-K+ pump

Explanation:

Active transport by the Na+-K+ pump

Maintenance (and restoration) of the resting ion concentrations depends on the Na+-K+ pump. Once gated ion channels are closed, the combined action of the pump and ion leakage (particularly that of K+) establishes a resting membrane potential in a typical neuron of around âˆ'70 mV.

5 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
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