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2 years ago

Describe two techniques used to measure the PH of a solution

Chemistry

2 answers:

natali 33 [55]2 years ago

8 0

Answer:

ph paper, or a ph meter

Explanation:

lord [1]2 years ago

3 0

Answer:

Measure H+ concentration or measure OH- concentration

Explanation:

Since pH is basically the number of H+ ions in a solution, that's what you're trying to find. (it's actually -log of H+, which is irrelevant since this is a conceptual question). You can do this with a variety of chemicals.

And p,OH is always 14-pH. So you can also use the number of OH- ions to find p,OH, which can find pH.

hope this helps

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Answer:

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2 years ago

What causes the Moon to be illuminated?

alexgriva [62]

Answer:

reflection

Explanation:

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8 0

1 year ago

Is classified as a substance <br><br> A. Elements and compound <br><br> B.mixtures

Vika [28.1K]

Pretty sure it’s Mixture if I’m not wrong

8 0

2 years ago

Brainliest for an answer!

Degger [83]

The volume of H₂ : = 15.2208 L

<h3>Further explanation</h3>

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

5 0

2 years ago

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!

3 0

3 years ago