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Sphinxa [80]
3 years ago
10

What is the most common ion for cesium ? Cs+1 Cs Cs-1 Cs-2

Chemistry
1 answer:
alisha [4.7K]3 years ago
5 0
Cs+1
The only common oxidation state is +1.
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When an electron in a quantum system drops from a higher energy level to a lower one, the system __
icang [17]

When an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>

<u />

The energy of the electron drops when it transitions levels, as well as the atom releases photons. The emission of the photon occurs as the electron transitions from an energy state to a lower state. The photon energy represents precisely the energy that would be lost when an electron moves to a level with less energy.

When such an excited electron transitions from one energy level to another, this could emit a photon. The energy drop would be equivalent to the power of the photon that is released. In electron volts, the energy of an electron, as well as its associated photon (emitted or absorbed) has been stated.

Therefore, when an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>

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1 year ago
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krok68 [10]

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7 0
3 years ago
2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0
2 years ago
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