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Likurg_2 [28]
2 years ago
7

Table salt contains 39.33 g of sodium per 100 g of salt. The U.S. Food and Drug Administration (FDA) recommends that adults cons

ume less than 2.40 g of sodium per day. A particular snack mix contains 1.30g of salt per 100 g of the mix.What mass of the snack mix can you consume and still be within the FDA limit?
Chemistry
1 answer:
Naya [18.7K]2 years ago
7 0

The  individual can consume less than 184.6 g of the snack mix and still be within the FDA limit of salt consumption.

<h3>What is the mass of snack that can be consumed within the limit of sodium intake?</h3>

The mass of the snack mix that the individual can consume and still be within the FDA limit is calculated as follows:

U.S. Food and Drug Administration (FDA) recommends of sodium intake = less than 2.40 g of sodium per day.

Amount of salt in 100 g of snack mix = 1.30 g

Mass of snack that will contain 2.40 g of sodium = 2.40 * 100g/1.30 = 184.6 g of snack mix

Therefore, the individual can consume less than 184.6 of the snack mix and still be within the FDA limit of salt consumption.

In conclusion, the FDA recommends that an individual take in less than 2.40 g of sodium per day from their diet.

Learn more about salts at: brainly.com/question/23463868

#SPJ1

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Harrizon [31]

Answer:

The answer to your question is None of your answers is correct, maybe the data are wrong.

Explanation:

Data

Concentration 1 = C1 = 1 M

Volume 2 = 5 ml

Concentration 2 = 0.05 M

Volume 1 = x

To solve this problem use the dilution formula

             Concentration 1 x Volume 1 = Concentration 2 x Volume 2

Solve for Volume 1

              Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1

Substitution

              Volume 1 = (0.05 x 5) / 1

Simplification

              Volume 1 = 0.25/1

Result

               Volume 1 = 0.25 ml

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4 years ago
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Answer:

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Explanation:

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2 years ago
Calculate the molality for each of the following solutions. Then, calculate the freezing-point depression ΔTF = iKFcm produced b
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Answer:

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b) Cm= 0.21 m ; ΔTf= 0.79ºC

Explanation:

In order to solve the problems, we have to remember that the molality (m) of a solution is equal to moles of solute in 1 kg of solvent.

m= mol solute/kg solvent

a) In this case we have molarity, which is moles of solute in 1 liter of solution. We have to know how many kg of solvent (water) we have in 1 L of solution.

3.2 M NaCl= 3.2 mol NaCl/ 1 L solution

1 L solution= 1000 ml solution x 1.00 g/ml= 1000 g

A solution is composed by solute (NaCl) + solvent, so:

1000 g solution = g NaCl + g solvent

g NaCl= 3.2 mol NaCl x 58.44 g/mol= 187 g NaCl

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ΔTf= i x KF x Cm= 2 x 1.86ºC/m x 3.9 m= 14.51ºC

b) In this case we have 24 g of solute in 1.5 L of solvent. We have to convert the liters of solvent to kg, and to convert the mass of solute to mol by using the molecular weight of KCl (74.55 g/mol):

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KCl is an electrolyte and when it dissolves in water, it dissociates in 2 ions: K⁺ and Cl⁻. For this, van't Hoff factor (i) is equal to 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC x 0.21 m= 0.79ºC

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Hi!

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