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olga55 [171]
2 years ago
13

A food handler puts a food thermometer into a pan of green bean casserole that is being hot-held on the serving line. The thermo

meter reads 146° F. Can the casserole be safely served to customers?
Chemistry
1 answer:
Sidana [21]2 years ago
4 0

Answer:

Yes , the food can be served.

Explanation:

Given the food here is served at 146°F

  • According to FDA(US food and drug administration) Hot foods should be kept at an internal temperature of 140°F or warmer.
  • This is the temperature recquired to maintain the bacteria without spoiling the food .
  • Dishes related to eggs like quiches or soufflés should be served at a minimum of 165°F

Also given that the temperature of the food being served is 146°F which is greater than 140°F(which is the minimum value) Therefore the food can be served.

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Balance the following oxidation-reduction reaction: Fe(s)+Na+(aq)→Fe2+(aq)+Na(s) Express the coefficients as integers separated
aliina [53]

Answer:

The answer to your question is: 1, 2, 1, 2

Explanation:

                       1 Fe(s)  + 2 Na⁺(aq)  → 1 Fe²⁺(aq)  + 2 Na(s)

                             Fe⁰   -   2e⁻       ⇒           Fe⁺²        Oxidases

                             Na⁺   +  1 e⁻       ⇒           Na⁰         Reduces

                      1 x ( 1 Fe⁰      ⇒         1 Fe⁺²)      Interchange number of

                      2 x ( 2Na⁺       ⇒       2 Na⁰ )      electrons

6 0
2 years ago
If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S​

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S​ = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S​

1 g of NH₃ will produce \frac{68}{34} g of (NH₄)₂S​

25g of NH₃ will produce \frac{65}{34} X 25 g of (NH₄)₂S​

                                     = 47.8 g of (NH₄)₂S​

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

4 0
2 years ago
A precipitate of zinc hydroxide can be formed using the reaction below.
worty [1.4K]

Answer:

Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

Explanation:

Step 1: Data given

Number of moles ZnCl2 = 0.36 moles

Number of moles KOH  = 0.54 moles

Step 2: The balanced equations

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

Step 3: Calculate the limiting reactant.

KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles

There will remain 0.36 - 0.27 = 0.09 moles.

Step 4: The products

There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)

Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol

Option B is correct

Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

7 0
2 years ago
How many moles of lead, Pb, are in 1.50 x 1012 atoms of lead?
SpyIntel [72]

Answer:

2.49*10⁻¹² mol

Explanation:

Use Avogadro's Number for this equation (6.022*10²³).  Divide Avogrado's by the number of atoms you have to find moles.  You are answer should be 2.49*10⁻¹² mol.

7 0
2 years ago
Read 2 more answers
Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C
timurjin [86]

Answer:

The specie which is oxidized is:- CrO_2^-

The specie which is reduced is:- ClO^-

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

For the given chemical reaction:

2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-

The half cell reactions for the above reaction follows:

Oxidation half reaction:  CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-

Reduction half reaction:  2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-

Thus, the specie which is oxidized is:- CrO_2^-

The specie which is reduced is:- ClO^-

8 0
3 years ago
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