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3 years ago

2. List and elaborate on at least two limitations of the 24-hour recall as a

Chemistry

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:

The correct answer is - may not be typical, and participant burden.

Explanation:

The 24-hour recall is nothing but a retrospective method of diet assessment. In this method, an individual is interviewed about his or her diet consumption during the last 24 hours.

The disadvantages or limitations of this method include the inability of a single day's intake to describe the typical diet, multiple recalls to intake, cost and administration time; participant burden, have to recall to reliably estimate usual intake.

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The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.

Nataliya [291]

a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = $\frac{mass of element}{mass of compound}$ (1)

The molecular formula of cisplatin is $Pt(NH_3)_2Cl_2$ .

The atomic weight of the elements in cisplatin is:

Platinum, $Pt = 195.084 u$

Nitrogen, $N = 14.0067 u$

Hydrogen, $H = 1 u$

Chlorine, $Cl = 35.453 u$

The molar mass of $Pt(NH_3)_2Cl_2$ = $195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453)$ = $300.00 g/mol$

The mass of each element calculated using formula (1):

- Platinum, $Pt$ %

$\frac{195.084}{300.00} \times 100 = 65.23$ %.

- Nitrogen, $N$ %

$\frac{2\times 14.0067}{300.00} \times 100 = 9.34$ %

- Hydrogen, $H$ %

$\frac{6\times 1}{300.00} \times 100 = 2.0$ %

- Chlorine, $Cl$ %

$\frac{2\times 35.453}{300.00} \times 100 = 23.63$ %

b. The given reaction of cisplatin is:

$K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)$

According to the balanced reaction, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ .

Now, calculating the number of moles of $K_2PtCl_4$ in 100.0 g.

Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

Since, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ . Therefore, mass of cisplatin is:

$0.241 mole\times 300 g/mol = 72.3 g$

For mass of $KCl$ :

Molar mass of $KCl$ = $39.0983 + 35.453 = 74.55 g/mol$

Since, 1 mole of $K_2PtCl_4$ gives 2 mole of $KCl$ . Therefore, mass of $KCl$ is:

$0.241 mole\times 74.55 g/mol\times 2 = 35.93 g$

5 0

2 years ago

You have an aqueous solution with a ph of 8.0. You add sodium chloride to a concentration of 1 gram per 100 milliliters. What ha

zysi [14]

Answer:

what is the ph of sodium chloride

4 0

3 years ago

Read 2 more answers

H3C6H507 + H2O + H3O+ + H2C6H507<br> acid <br> base

Aleonysh [2.5K]

Explanation:

an acid will give away a proton and become a conjugate base.

A base will accept a proton and become a conjugate acid.

3 0

3 years ago

If I have 7.5 L of a gas in a piston at a pressure of 1.75 atm and compress the gas until its volume is 3.7 L, what will the new

Sergio [31]

Answer:

3.55atm

Explanation:

We will apply Boyle's law formula in solving this problem.

P1V1 = P2V2

And with values given in the question

P1=initial pressure of gas = 1.75atm

V1=initial volume of gas =7.5L

P2=final pressure of gas inside new piston in atm

V2=final volume of gas = 3.7L

We need to find the final pressure

From the equation, P1V1 = P2V2,

We make P2 subject

P2 = (P1V1) / V2

P2 = (1.75×7.5)/3.7

P2=3.55atm

Therefore, the new pressure inside the piston is 3.55atm

7 0

3 years ago

Describe the double circulation system of the human body

Alika [10]

Answer:

we have two loops in our body in which blood circulates. One is oxygenated, meaning oxygen rich, and the other is deoxygenated, which means it has little to no oxygen, but a lot of carbon dioxide.

4 0

3 years ago