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Molodets [167]
3 years ago
7

0.0400 M solution of HClO2, pH = 1.80 solve for Ka

Chemistry
1 answer:
umka2103 [35]3 years ago
6 0
Ka for weak acids is ---> Ka= (X)²/ (initial- X)

initial= 0.0400 M

to find X, you can find the H concentration using pH

[H+]= 10^-pH
[H+]= 10^-1.80= 0.016 which is X

ka = (0.016)²/ (0.0400 - 0.016) = 0.011
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AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the
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The question is incomplete, the complete question is;

AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —

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aluminum tribromide

Answer:

aluminum bromide

Explanation:

Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.

We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.

Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.

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3 years ago
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son4ous [18]
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timofeeve [1]

Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

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\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

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4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

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The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

0.80 \ mol \ F

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

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