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3 years ago

What must a food handler use to determine the concentration of sanitizing solution?

Chemistry

2 answers:

Marina86 [1]3 years ago

8 0

Answer:

He must test the solution with a sanitizer kit.

Explanation:

The concentration of sanitizing solution must be tested with a sanitizing kit. It is composed of disinfectant test strips that measure the sanitizing solution concentration. A very important part of the daily operation of a food establishment is the proper disinfection of the equipment. Generally, chlorine is used to disinfect multipurpose utensils, food contact surfaces and other food preparation items. Over time, the concentration of chlorine decreases due to volatile elements.

Have a nice day!

sattari [20]3 years ago

3 0

I believe a food handler must use a test kit to determine the concentration of sanitizing solution. Sanitizing reduces pathogens on a surface to safe levels. The sanitizer are the agents that are used for sanitizing ; their effectiveness is determined by the factors such as concentration, contact time, pH, temperature, and also water hardness.

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2 years ago

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3 years ago

Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?

Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

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= 6/2 x 0.134

= 0.402

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2 years ago

Marie discovers boxes of elements in a storage room. The boxes do not say which elements they contain, but they do have informat

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3 years ago

Read 2 more answers

The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

tankabanditka [31]

Explanation:

In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.

Hence, the half-cell reaction taking place at anode and cathode will be as follows.

At anode (Oxidation) : $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode (Reduction) : $Ag^{+}(aq) + 1e^{-} \rightarrow Ag(s)$

So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.

$3Ag^{+}(aq) + 3e^{-} \rightarrow 3Ag(s)$ ........ (2)

Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.

$Cr(s) 3Ag^{+}(aq) \rightarrow Cr^{3+}(aq) + 3Ag(s)$

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3 years ago