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3 years ago

How can scientists detect blood doping?

Chemistry

2 answers:

aev [14]3 years ago

6 0

Answer:

There are typically three ways that it is accomplished: use of erythropoietin (EPO) or synthetic oxygen carriers and blood transfusions. While transfusions of large volumes of blood or use of EPO can be detected, microdosing EPO or transfusing smaller volumes of packed red blood cells is much harder to detect.

vesna_86 [32]3 years ago

6 0

Answer:

Explanation:

Blood doping via homologous transfusion can be detected by testing. The tests were used at the 2004 Summer Olympic Games in Athens, Greece. EPO injections. Blood and urine tests can detect the presence of synthetic EPO

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What is the difference between a compound and a molecule?

Sergio [31]

A molecule is formed when two or more atoms joined together chemically. A compound is a molecule that contains at least two different elements. All compounds are molecules. Hope I helped

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3 years ago

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Explain how to use an acid-base indicator to determine pH

Annette [7]

Acid-base indicator changes color based on pH.

drop some in a solution n watch the color changes. different indicators show different colors at different pH. they usually have standard colors for comparison.

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3 years ago

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An element has an atomic mass number of 16 and an atomic number of 7

Kitty [74]

Answer: Nitrogen

Explanation: Nitrogen has an atomic number of 7 and an atomic mass of about 14 u

7 0

3 years ago

The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.

avanturin [10]

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

6 0

3 years ago

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago