How much heat is absorbed when 52.39 H2O(l) at 100°C and 101.3 kPa is converted to steam at 100°C? (The molar heat of vaporization of water is40.7 k/mol.)
2.09 x 1020
O 2.31% 10110
O 1.18 x 102 103
O 1.11% 1021)

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.
• Efficacy of exercise interventions in patients with advanced cancer: A systematic review. Archives of Physical Medicine and Rehabilitation, May 2018.

• 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.

How would you market your services to clients that have cancer?

The benefits of exercise against cancer are innumerable: it helps prevent it, reduces the side effects of chemotherapy and radiotherapy, decreases cancer recurrence, improves vital energy, mobility and balance and reduces fatigue, maintains muscle mass, improves self-esteem and sleep quality, decreases the level of anxiety, depression and stress.

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.

Research the benefits and risks of exercise and youth.

The benefits of physical activity and sports in young people imply a better physical condition, but also plays a fundamental role from the psychological and social. Every healthy habit is best incorporated from childhood, so that it becomes natural and everyday and improves the quality of life of our future adults.

The benefits of physical activity in youth are several:

• Better cardiorespiratory function and greater muscular strength

• Fat reduction, children and young people who perform physical activity have lower body fat.
• Decreased risk of subsequent cardiovascular and metabolic diseases such as high blood pressure, diabetes, high cholesterol.
• Better bone health, because the growing bones are strengthened.
• Fewer symptoms of depression since they do not get bored, find motivations and social relationships.

Children and young people should perform daily physical activities in the form of commuting, games, recreational activities, physical education, programmed exercises and sports, in the context of school and clubs, if possible integrating other family members.

Final answer:

To neutralize the KOH solution, we need 61.4 mL of 1.33 mol L−1 H2SO4(aq).

Explanation:

To find the volume of the H2SO4 solution needed to neutralize the KOH solution, we can use the equation:

Mole of H2SO4 = Molarity of KOH x Volume of KOH

First, calculate the moles of KOH:
Moles of KOH = Molarity of KOH x Volume of KOH = 0.830 mol/L x (49.3 mL / 1000 mL) = 0.04089 mol

Since H2SO4 is a diprotic acid and KOH is a strong base, the reaction will be:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O

Therefore, the ratio between the moles of H2SO4 and KOH is 1:2. This means that twice the moles of KOH will be needed to neutralize the H2SO4. Calculate the moles of H2SO4 needed:
Moles of H2SO4 needed = 2 x Moles of KOH

= 2 x 0.04089 mol

= 0.08178 mol

Finally, calculate the volume of the H2SO4 solution needed:
Volume of H2SO4 = Moles of H2SO4 / Molarity of H2SO4 = 0.08178 mol / 1.33 mol/L

= 0.0614 L

= 61.4 mL

Learn more about Neutralization here:

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Your answer is 160 grams I hope this helps

Answer:

1. Al is consumed first and CuSO₄ remains left.

2. The grams of ferric chloride that forms is 14.5 g.

3. The percent yield is 63.1%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.

Explanation:

1. The reaction is:

2Al + 3CuSO₄ = Al₂(SO₄)₃ + 3Cu

The number of moles of Al is less than the number of moles of CuSO₄. Therefore, Al is the limiting reagent and CuSO₄ is the excess reagent. This means that Al is consumed first and CuSO₄ remains left.

2. The reaction is:

2Fe + 3Cl₂ = 2FeCl₃

The number of moles of Fe is:

The number of moles of Cl₂ is:

We know that 2 moles of Fe react with 3 moles of Cl₂, thus:

2 moles Fe---------------3 moles Cl₂

0.0895 moles Fe-------X moles Cl₂

Clearing X:

It needs 0.134 moles of Cl₂ but it only has 0.211 moles, thus, Cl₂ is the excess reagent. Fe is the limiting reagent.

2 moles Fe-----------2 moles FeCl₃

0.0895 moles Fe------X moles FeCl₃

Clearing X:

3. The actual yield of FeCl₃ is 9.15 g, the theoritical yield is 14.5 g, thus, ther percent yield is:

%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.

• 4.289

Explanation:

4.289 x 10^0 can be written as a regular number by moving the decimal point to the right or left based on the exponent value. Since the exponent is 0, the decimal point does not need to be moved. Therefore, the regular number form of 4.289 x 10^0 is simply 4.289.