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Bogdan [553]
1 year ago
6

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

then mixed with 75.0 mL of a 0.800 M aqueous solution of AlBr3. Calculate the concentration (M) of Na+ and Br− in the final solution.
Chemistry
1 answer:
Colt1911 [192]1 year ago
7 0

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

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Answer:

Gd → Gd⁺ + 1e⁻, Gd⁺ → Gd⁺² + 1e⁻, Gd⁺² → Gd⁺³ + 1e⁻

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Fisrt ionization:

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Second ionization:

Gd⁺ → Gd⁺² + 1e⁻

Third ionization:

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3 0
2 years ago
Write a balanced equation for the transmutation that occurs when a scandium-48 nucleus undergoes beta decay.
Tju [1.3M]

Answer:

A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.

\rm ^{48}_{21}Sc \to ^{48}_{22}Ti + ^{\phantom{1}\,0}_{-1}e^{-} + \bar{\mathnormal{v}}_e.

Explanation:

There are two types of beta decay modes: beta-minus and beta-plus.

In both decay modes, the mass number of the nucleus stays the same.

However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.

Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.

Look up the atomic number and relative atomic mass for the element scandium.

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  • The relative atomic mass of \rm Sc is approximately 45.0.

This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.

Each element (e.g, \rm Sc) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average  across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.

The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.

The atomic number of the nucleus will increase by 1. 21 + 1 = 22. That corresponds to titanium. The mass number stays the same at 48. Hence the daughter nucleus would be titanium-48. Note that two other particles: one electron and one electron \rm e^{-} and one antineutrino \bar{v}_{\text{e}} (note the bar.) The neutrino helps balance the lepton number of this reaction.

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1 year ago
The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?
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3 0
2 years ago
Analysis of the water content of a lake found in the desert showed that it contained 16.6 percent chloride ion, and had a densit
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Answer : The molarity of the chloride ion in the water is, 5.75 M

Explanation :

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Now we have to calculate the molarity of chloride ion.

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Formula used :

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Now put all the given values in this formula, we get:

\text{Molarity}=\frac{16.6g\times 1000}{35.5g/mole\times 81.3mL}=5.75mole/L=5.75M

Thus, the molarity of the chloride ion in the water is, 5.75 M

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