When selecting small wares to use in the kitchen, a manager must be sure the small wares are durable, easy to clean, safe, and ?

Chemistry

1 answer:

Ilia_Sergeevich [38]4 years ago

8 0

Answer:

Non-reactors

Explanation:

Kitchen wares should preferably be non-reactors to prevent them from reactiong with food.

Reactions between food substances and wares lead to production of untested and unsecure reaction results.

The production of this predictably hazardous results may result to food poisoning thereby threatening human life.

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Ephedrine, a central nervous system stimulant, is used in nasalsprays as a decongestant. this compound is a weak organic base: {

sashaice [31]

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-?

b) Calculate Kb for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH⁻] = 10∧(-2,67) = 0,00213 M.
[OH⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.
[C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.
Kb = [OH⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].
Kb = (0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

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Which salt is formed by the reaction between hydrochloric acid and sodium hydroxide, write with chemical equation. .

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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$