CHEMISTRY COLLEGE

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine produces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.

Answers

Answer 1

Answer:

The molar mass of the compound is found by finding the empirical and

molecular formula of the compound.

The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Reasons:

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ produced = $(18.13)/(44.01)$ ≈ 0.412 moles

Number of moles of produced C = 0.412 moles

Mass of C = 12 × 0.412 = 4.944 g

Molar mass of H₂O = 18.015 g/mol

Moles of H₂O produced = $(4.639)/(18.015 )$ = 0.2575 moles

Moles of produced H = 2 × 0.2575 = 0.515 moles

Mass of H = 1.00784 × 0.515 ≈ 0.519 g

Molar mass of N₂ = 28.0134 g/mol

Moles of N produced = 2 × $(2.885)/(28.0134 )$ = 2 × 0.103 = 0.206 moles

Mass of oxygen = 10 - 4.944 - 2.885 - 0.519 = 1.652

Moles of oxygen, O = $(1.652 \, g)/(16 \, g/mol)$ ≈ 0.103 moles

Therefore, we get;

Number of moles of produced C = 0.412 moles

Number of moles of produced H = 0.515 moles

Number of moles of oxygen, O ≈ 0.103 moles

Number of moles of N produced = 0.206 moles

Dividing by 0.103 gives;

Mole ratio of C = $(0.412)/(0.103) = 4$

Mole ratio of H = $(0.515)/(0.103) = 5$

Mole ratio of O = 1

Mole ratio of N = $(0.206)/(0.103) = 2$

The empirical formula of the compound is therefore; C₄H₅N₂O

The general molecular formula is of the form (C₄H₅N₂O)ₙ

Molar mass of the compound is between 150 g/mol and 210 g/mol (given)

The molar mass of C₄H₅N₂O = 4×12 + 5×1.00784 + 2×14 + 16 ≈ 97

The molar mass of C₄H₅N₂O ≈ 97 g/mol

Molar mass of the compound is between 150 and 210 g/mol, therefore, n in

(C₄H₅N₂O)ₙ = 2, which gives;

The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

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Answer 2

Answer:

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

$C=18.13gCO_2*(12gC)/(44gCO_2) =4.945gC\nH=4.639gH_2O*(2.016gH)/(18.0152gH_2O)=0.519gH\nN=2.885gN_2\nO=10.0g-4.945g-0.519g-2.885g=1.651gO$

Next, the corresponding moles:

$C=4.945gC*(1molC)/(12gC)=0.412mol\nH=0.519gH*(1molH)/(1gH)=0.519mol\nN=2.885gN*(1molN)/(14gN)=0.206molN\nO=1.648gO*(1molO)/(16gO) =0.103molO$

Then, each element's subscripts is found to be:

$C=(0.412)/(0.103)=4\nH=(0.519)/(0.103)=5\nN=(0.206)/(0.103) =2\nO=(0.103)/(0.103)=1$

Therefore, the empirical formula is:

$C_4H_5N_2O$

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

$C_8H_10N_4O_2$

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

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Mg+
O
Cl-
Na+

Answers

Answer:

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Explanation:

Answer:

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When the following redox equation is balanced with smallest whole number coefficients, the coefficient for Sn(OH)3– will be _____. Bi(OH)3(s) + Sn(OH)3–(aq) → Sn(OH)62–(aq) + Bi(s) (basic solution)

Answers

Answer:

The coefficient of $Sn(OH)_3^(-)$ is 3 in the balanced redox reaction.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom loses 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.

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For the given chemical reaction:

$Bi(OH)_3+Sn(OH)_3^(-)\rightarrow Sn(OH)6^(2-)+Bi$

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To balance the oxidation half reaction must be multiplied by 3 and reduction half reaction must be multiplied by 2 thus, the balanced equation is:-

$3Sn(OH)_3^(-)+3OH^-+2Bi(OH)_3\rightarrow 3Sn(OH)6^(2-)+2Bi$

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Search online for "free medline." You will probably find several websites that offer this feature. Search for research abstracts on "cancer and exercise." Review at least six abstracts (articles no more than five years old). Based on the conclusions of these studies, how beneficial is regular exercise for cancer patients? How would you market your services to clients that have cancer? Be sure to cite your work.Research the benefits and risks of exercise and youth. List at least five resources (resources no more than five years old) and summarize the research findings in your own words. Is resistance training safe, effective, and beneficial for young people? Why or why not?

Answers

Answer:

There are currently a variety of advanced medical treatment screening programs for certain types of cancer that have resulted in more people having a better chance of healing or living longer.

Explanation:

Exercise helps cancer survivors cope with and recover from treatment; exercise may improve the health of long term cancer survivors and extend survival. Physical exercise will benefit throughout the spectrum of cancer. However, an understanding of the amount, type and intensity of exercise needed has not been fully elucidated. There is sufficient evidence to promote exercise in cancer survivors following careful assessment and tailoring on exercise prescription.

"The field of oncology will benefit from understanding the importance of physical activity both for primary prevention as well as in helping cancer survivors cope with and recover from treatments, improve the health of long term cancer survivors and possibly even reduce the risk of recurrence and extend survival after a cancer diagnosis" (P. Rajarajeswaran, R. Vishnupriya)

Additional studies will be needed to more firmly establish physical activity benefits to cancer survivors.

R. Segal, MD*, C. Zwaal, MSc†, E. Green, RN‡, J.R. Tomasone, PhD§, A. Loblaw, MD MSc‖, T. Petrella, MD, Exercise for people with cancer: a clinical practice guideline. 2017. Canadian Cancer Research Journal.

A systematic review and meta-analysis of the safety, feasibility and effect of exercise in women with stage II+ breast cancer. Archives of Physical Medicine and Rehabilitation, May 2018.
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McNeely ML. Exercise as a promising intervention in head & neck cancer patients. Indian J Med Res.
P. Rajarajeswaran, R. Vishnupriya. Exercise in cancer. College of Physiotherapy, Mother Theresa Post Graduate and Research Institute of Health Sciences, India.

Exercise is key both in the prevention and treatment of cancer, since it improves the quality and life expectancy of patients.

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No one doubts the importance of physical activity, exercise and sport in global health, in the prevention and even in the treatment of numerous diseases. Among these diseases is cancer. There are more than 10,000 scientific publications that have studied the links between exercise and cancer and almost all of them with positive results regarding the prevention of numerous types of tumors, the decrease in cancer recurrence and the best prognosis of the latter if You exercise.

It is scientifically proven that properly prescribed physical exercise can be performed without risk during and after chemotherapy and radiotherapy treatments. However, it is necessary to adjust its intensity, duration, weekly frequency and type of exercise to the general condition of the patient. Physical exercise will improve the quality of life, fatigue and mood of the cancer patient being treated. It will also improve the prognosis of the disease, its quality of future life and its final life expectancy.

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Calculate the mass percent of oxygen in KMnO4.

Answers

Answer: 40.496%

Hope this helps! (:

What is the total mass (amu) of carbon in each of the following molecules?(a)CH4
(b)CHCL3
(c)C12H10O16
(d)CH3CH2CH2CH2CH3

Answers

The mass of carbon in $\rm CH_4$ is 12.007 amu, $\rm CHCl_3$ is 12.007, $\rm C_1_2H_1_0O_1_6$ is 144.084 amu, and $\rm CH_3CH_2CH_2CH_2CH_3$ is 60.035 amu.

What is the mass of one Carbon atom?

The mass has been given as the sum of the atomic mass unit in the compound. The mass of 1 atom of carbon is 12.007 amu.

The mass of carbon in the following compounds is given as:

$\rm CH_4$

The number of Carbon units = 1

The mass of carbon in compound = 12.007 amu

$\rm CHCl_3$

The number of Carbon units = 1

The mass of carbon in the compound = 12.007 amu

$\rm C_1_2H_1_0O_1_6$

The number of carbon units =12

The mass of carbon in the compound:

$\rm 12\;*\;12.007\;amu\n=144.084\;amu$

$\rm CH_3CH_2CH_2CH_2CH_3$

The number of carbon units = 5

The mass of carbon in the compound:

$\rm 5\;*\;12.007\;amu\n=60.035\;amu$

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Answer:

The atomic mass of carbon (C) is 12.0107 amu, so if you want to calculate the total mass in each molecule, you just need to multiply the number of carbon atoms in the substance by 12.017. In (a) there is one atom of C, (b) have also one atom of C, (c) have 12 atoms of C, and (d) have five atoms of C. Thus, the total mass (amu) of carbon is:

(a) 12.017 amu

(b) 12.017 amu

(d) 60.085 amu

Calculate the molarity of a solution that contains 0.50 g of NaCl dissolved in 100mL of solution?

Answers

The molarity is an important method which is used to calculate the concentration of a solution. The molarity of a solution that contains 0.50 g of NaCl dissolved in 100mL of solution is 0.085 M.

What is molarity?

The molarity of a solution is defined as the number of moles of the solute present per litre of the solution. It is an most important method to calculate the concentration of a binary solution. It is represented as 'M'.

The equation used to calculate the molarity is:

Molarity = Number of moles of the solute / Volume of the solution in litres

1L = 1000 mL

100 mL = 0.1 L

Number of moles (n) = Given mass / Molar mass

n = 0.50 / 58.44 = 0.008

Molarity = 0.0085 / 0.1 = 0.085 M

Thus the molarity of the solution is 0.085 M.

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#SPJ3

The answer is 0.085470.. mol/dm^3. I might have made some mistakes but I have included the working out so do try it yourself to make sure it’s right!

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