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Tcecarenko [31]
3 years ago
6

A student is given 50.0mL of a solution of  Na2CO3  of unknown concentration. To determine the concentration of the solution, th

e student mixes the solution with excess 1.0MCa(NO3)2(aq) , causing a precipitate to form. The balanced equation for the reaction is shown below.
Na2CO3(aq)+Ca(NO3)2(aq)→2NaNO3(aq)+CaCO3(s)

(a) Write the net ionic equation for the reaction that occurs when the solutions of  Na2CO3  and  Ca(NO3)2  are mixed.
Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0

Answer:

Ca(aq)⁺²  + CO₃⁻²(aq) → CaCO₃(s)

Explanation:

Breaking down the equation into ionic form gives:

2Na⁺(aq) + CO₃⁻²(aq) + Ca⁺²(aq) + 2NO₃⁻¹ (aq)  →  2Na⁺(aq) + 2NO₃⁻¹(aq)  + CaCO₃(s)

Eliminating all the same ionic states on both sides of the equation gives following final equation

Ca(aq)⁺²  + CO₃⁻²(aq) → CaCO₃(s)

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the symbol for xenon (xe) would be a part of the noble gas notation for the element antimony. cesium.
trapecia [35]

The symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.

For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.

A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.

The actual electronic configuration of antimony is as follow:

[Kr] 4d10 5s2 5p3

B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.

The actual electronic configuration of Cesium is as follow:

[Xe] 6s1

Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.

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How many particles are in a 151 g sample of Li2O?
neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

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