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Orlov [11]
3 years ago
7

What is 6.25 in scientific notation?

Chemistry
2 answers:
Contact [7]3 years ago
7 0
I think it’s 6.25 x 10^2
charle [14.2K]3 years ago
4 0
It is 6.25x10^0 I believe :)
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Is anyone good at chemistry if so can someone help me please ?<br><br> (NO LINKS)
Mkey [24]

This requires familiarity with the different theories (or concepts) of acids and bases.

On the Arrhenius concept, an acid is a substance that produces an H⁺ ion in water such that the H⁺ concentration increases, and a base is a substance that produces an OH⁻ ion in water such that the OH⁻ concentration increases.

On the Brønsted–Lowry concept, an acid is a substance that donates a proton (which is basically an H⁺ ion) in a solvent, and a base is a substance that accepts a proton in a solvent.

On the Lewis concept, an acid is a substance that accepts an electron pair in a solvent, and a base is a substance that donates an electron pair in a solvent.

The concepts become progressively broader, i.e., the Arrhenius concept is the most restrictive and the Lewis concept is the least restrictive. As a corollary, an Arrhenius acid or base is also both a Brønsted–Lowry acid or base and a Lewis acid or base, respectively; a Brønsted–Lowry acid or base is not necessarily an Arrhenius acid or base, but an Arrhenius acid or base is also a Lewis acid or base, respectively. And finally, a Lewis acid or base may not necessarily be either an Arrhenius or a Brønsted–Lowry acid or base.

So, with the above concepts in mind, we can match the statements in column A with the type of acid or base in column B:

\begin{center}\begin{tabular}{ c c } 1 & Bronsted Lowry acid \\  2 & Bronsted Lowry base \\   3 & Arrhenius acid \\ 4 & Arrhenius base \\ 5 & Lewis base \\ 6 & Lewis acid\end{tabular}\end{center}

6 0
3 years ago
What is the average atomic mass of hafnium given the following abundance information on the isotopes?
Goryan [66]

Answer:

178.55

Explanation:

176 × 0.05+ 177 × 0.19 + 178 × 0.27 + 179 × 0.14 + 180 × 0.35 = 178.55

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by statistical analyses, especially by determining the p-value

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