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4 years ago

6

If the pH value was 3.6, what is the [H+]? 2.5 x 10-4 4.0 x 103

2 answers:

KengaRu [80]4 years ago

6 0

If the ph value was3.6,H+is2.5x10-4

Soloha48 [4]4 years ago

4 0

<u>Answer:</u> The hydrogen ion concentration for the given pH value is $2.5\times 10^{-4}$

<u>Explanation:</u>

pH is defined as the negative logarithm of hydrogen ion or hydronium ion concentration. The equation used to represent pH of the solution follows:

$pH=-\log[H^+]$

We are given:

pH of the solution = 3.6

Putting values in above equation, we get:

$3.6=-\log[H^+]$

$[H^+]=2.5\times 10^{-4}$

Hence, the hydrogen ion concentration for the given pH value is $2.5\times 10^{-4}$

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About 30% of all fresh water is easily accessible by humans. <br> Where can this water be found?

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Answer:

lake and streams

Explanation:

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3 years ago

A chemist used 6.5 moles of water in this reaction. How many grams of water were used?

denpristay [2]

Answer:

$\boxed {\boxed {\sf 120 \ or \ 117 \ grams \ H_2O \ depending \ on \ significant \ figures }}$

Explanation:

We want to convert from moles of water to grams of water.

First, find the molar mass of water (H₂O) Look on the Periodic Table for the masses of hydrogen and oxygen.

Hydrogen (H): 1.008 g/mol
Oxygen (O): 15.999 g/mol

Next, add up the number of each element in water. The subscript of 2 comes after the H, so there are 2 moles of hydrogen.

2 Hydrogen: (1.008 g/mol*2) = 2.016 g/mol

Finally, add the molar mass of 2 hydrogen and 1 oxygen.

2.016 g/mol (2 Hydrogen) + 15.999 g/mol (1 oxygen)= 18.015 g/mol

Next, find the grams in 6.5 moles.

Use the molar mass we just found as a ratio.

$molar \ mass \ ratio: \frac{18.015 \ g \ H_2O}{1 \ mol \ H_2O}$

We want to find the grams in 6.5 moles. We can multiply the ratio above by 6.5

$6.5 \ mol \ H_2O * \frac{18.015 \ g \ H_2O}{1 \ mol \ H_2O}$

Multiply. Note that the moles of H₂O will cancel each other out.

$6.5 * \frac{18.015 \ g \ H_2O}{1}$

$6.5 * {18.015 \ g \ H_2O}$

$117.0975 \ g \ H_2O$

If we want to round to the technically correct significant figures, it would be 2 sig figs. The original measurement, 6.5, has 2 (6 and 5).

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3 years ago

Given the two half-reactions, what must be done in the next step before the reaction can be balanced? Au3+ --> Au I- --> I

Artyom0805 [142]

Answer:

The balance equation is

$2Au^{+3} + 6I^{-}$ → $2Au + 3I_{2}$

Explanation:

first, we have to make sure that the atoms are balanced

$Au^{+3}$ → $Au^{}$

$2I^{-}$ → $I_{2}$

then we proceed to balance charges of each half-reaction

$Au^{+3} + 3e^{-}$ → $Au^{}$

$2I^{-}$ → $I_{2}$ $+ 2e^{-}$

Now we multiply the half-reactions to match the number of electrons in each one

( $Au^{+3} + 3e^{-}$ → $Au^{}$ )x2

( $2I^{-}$ → $I_{2}$ $+ 2e^{-}$ )x3

and now we do the sum of the half-reactions

2 $Au^{+3} + 6e^{-}$ → 2 $Au^{}$

$6I^{-}$ → 3 $I_{2}$ $+ 6e^{-}$

$2Au^{+3} + 6I^{-}$ → $2Au + 3I_{2}$

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