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ser-zykov [4K]
3 years ago
9

Determine the [OH⁻] concentration in a 0.344 M Ca(OH)₂ solution.

Chemistry
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

0.688M

Explanation:

Hello,

In this case, it is widely acknowledged that strong bases usually correspond to those formed with metals in groups IA and IIA which have relatively high activity and reactivity, therefore, when they are dissolved in water the following dissociation reaction occurs (for calcium hydroxide):

Ca(OH)_2\rightarrow Ca^{2+}+2OH^-

In such a way, for the same volume, we can compute the concentration of hydroxyl ions by simple stoichiometry (1:2 molar ratio):

0.344\frac{molCa(OH)_2}{L}*\frac{2molOH^-}{1molCa(OH)_2}  \\\\0.688\frac{mol OH^-}{L}

Or simply:

0.688M

Regards.

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For the reaction, Cl2 + 2KBr --> 2KCl + Br2, how many moles of potassium chloride, KCl, are produced from 102g of potassium b
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Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

m = 0.86 \times Molar \  mass \ of \  KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\  gram

Hence, this is the required solution.

5 0
3 years ago
You have a 5-liter container with 1.30 x 1024 molecules of ammonia gas (NH3) at STP.
bogdanovich [222]

Answer:

3). 1.30 × 10^(24) molecules

Explanation:

From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.

We can relate it to this question as;

V₁/n₁ = V₂/n₂

Where;

V₁ is initial volume

n₁ is initial number of molecules

V₂ is final volume

n₂ is final number of molecules

Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;

5/(1.30 x 10^(24)) = 5/n₂

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3 years ago
State general trend for metal properties as you go left to right across a period
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Electronegativity Trends

Electronegativity can be understood as a chemical property describing an atom's ability to attract and bind with electrons. Because electronegativity is a qualitative property, there is no standardized method for calculating electronegativity. However, the most common scale for quantifying electronegativity is the Pauling scale (Table A2), named after the chemist Linus Pauling. The numbers assigned by the Pauling scale are dimensionless due to the qualitative nature of electronegativity. Electronegativity values for each element can be found on certain periodic tables. An example is provided below.


From left to right across a period of elements, electronegativity increases. If the valence shell of an atom is less than half full, it requires less energy to lose an electron than to gain one. Conversely, if the valence shell is more than half full, it is easier to pull an electron into the valence shell than to donate one.

From top to bottom down a group, electronegativity decreases. This is because atomic number increases down a group, and thus there is an increased distance between the valence electrons and nucleus, or a greater atomic radius.

Important exceptions of the above rules include the noble gases, lanthanides, and actinides. The noble gases possess a complete valence shell and do not usually attract electrons. The lanthanides and actinides possess more complicated chemistry that does not generally follow any trends. Therefore, noble gases, lanthanides, and actinides do not have electronegativity values.

As for the transition metals, although they have electronegativity values, there is little variance among them across the period and up and down a group. This is because their metallic properties affect their ability to attract electrons as easily as the other elements.

According to these two general trends, the most electronegative element is fluorine, with 3.98 Pauling units.



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