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erastovalidia [21]
2 years ago
13

Daniel Harris has always enjoyed eating tuna fish. Unfortunately, a study of the mercury content of canned tuna in 2010 found th

at chunk white tuna contains 0.6 ppm Hg and chunk light tuna contains 0.14 ppm.2 The U.S. Environmental Protection Agency (EPA) recommends no more than 0.1 μg Hg/kg body weight per day. A person weighs 63 kg . How often may the person eat a can containing 6 ounces (16 oz=1 lb) of chunk white tuna so that they do not average more than 0.1 μg Hg/kg body weight per day?
А The person can consume one 6 oz can of chunk white tuna every_______ days.
If the person switches to chunk light tuna, how often can they eat one can?
The person can consume one 6 oz can of chunk light tuna every_______ days.
Chemistry
1 answer:
meriva2 years ago
4 0
That would be 3 days 6 oz of tuna
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Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a
geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

<u />

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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0
2 years ago
6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of
Dennis_Churaev [7]

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2

Moles of KHP = \frac{6.74 g}{204.2 g/mol}=0.0330 mol

According to reaction, 2 moles of KHP  with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

\frac{1}{2}\times 0.0330 mol=0.01650 mol of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

Molarity=\frac{Moles}{Volume(L)}

0.703 M=\frac{0.01650 M}{V}

V=\frac{0.01650 M}{0.703 M}=0.0235 mL

Volume of the calcium hydroxide solution used is 0.0235 mL.

4 0
2 years ago
Which graphic below shows the correct orientation of each water molecule when it is near the cl- ion in the aqueous solution?
andrew-mc [135]
The question is missing the graphics required to answer which I have attached as an image.

There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.

We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.

The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.

The correct representation is shown in graph 3 which shows all hydrogen atoms facing the chlorine anion.

8 0
2 years ago
How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?
VARVARA [1.3K]
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
4 0
2 years ago
Sodium nitrite (NaNO2)reacted with 2−iodooctane to give a mixture of two constitutionally isomeric compounds of molecular formul
finlep [7]

Answer:

The two constitutionally isomeric compounds that result from the reaction between sodium nitrite (NaNO2) and 2−Iodooctane are the following:

3 0
2 years ago
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