How many atoms of Bi are there in a 41.8 g sample?

Chemistry

1 answer:

Alekssandra [29.7K]2 years ago

3 0

Answer:

The number of particles in the 41.8g sample of Bismuth is 12.044 × 10²³

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A quantity of an ideal gas is compressed to half its initial volume. the process may be adiabatic, isothermal or isobaric. the g

JulsSmile [24]

the greatest amount of work is required if the process is adiabatic.The correct option is adiabatic.

The process in which heat is constant is called adiabatic process.

The The process in which temperature is constant is called isothermal process.

The process in which pressure is constant is called isobaric process.

The P-V diagram for adiabatic , isothermal and isobaric process is given below.

Work done in process = area encloses by P-V diagram axis . Since area under the curve is maximum for adiabatic process which is shown in the above diagram. So, work done by the gas will be maximum for adiabatic process.

learn more about adiabatic process.

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8 months ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.