How many water molecules are in a block of ice containing 1.50 mol of water (H2O)?

Answer:

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g) , Kp=1.112 at 150 K

denoting BC as BrCl , B as Br₂ , C as Cl₂, p as partial pressure , then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √(Kp*pB*pC)

replacing values

pBC = √(Kp*pB*pC) = √(1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Final answer:

To calculate the equilibrium partial pressure of BrCl, use the equilibrium constant expression and substitute the given partial pressures of Br2 and Cl2. The equilibrium partial pressure of BrCl is approximately 0.0375 atm.

Explanation:

To calculate the equilibrium partial pressure of BrCl, we need to use the equilibrium constant expression:

Kp = ([BrCl]^2) / ([Br2] * [Cl2])

Given that the equilibrium partial pressures of Br2 and Cl2 are 0.450 atm and 0.115 atm, respectively, we can substitute these values into the expression:

1.112 = ([BrCl]^2) / (0.450 * 0.115)

Simplifying the expression, we find that the equilibrium partial pressure of BrCl is approximately 0.0375 atm.

Learn more about Equilibrium partial pressure of BrCl here:

brainly.com/question/35557199

#SPJ12

Answer:

The new volume of the balloon is 46.1 L

Explanation:

Step 1: Data given

Initial volume of the balloon = 27.3 L

Initial pressure in the balloon = 738 mmHg = 0.97105 atm

Initial temperature in the balloon = 26.9 °C = 300.05 K

The pressure decreases to 375 mmHg = 0.493421 atm

The temperature lowers to -15.6 °C = 257.55 K

Step 2: Calculate the volume

P1*V1 / T1 = P2 *V2 / T2

⇒with P1 = the Initial pressure = 738 mmHg = 0.97105 atm

⇒with V1 = Initial volume of the balloon = 27.3 L

⇒with T1 = Initial temperature in the balloon = 26.9 °C = 300.05 K

⇒with P2 = the decreased pressure = 0.493421 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the lowered temperature = 257.55 K

0.97105 * 27.3 / 300.05 = 0.493421*V2 / 257.55

V2 = 46.1 L

The new volume of the balloon is 46.1 L

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

Recall, that the molar mass of the oxygen gas is 32 g/mol

Step 3; Convert the temperature to degree Kelvin

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

Hence, the volume of the oxygen gas is 17.5 L

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.

Answer: C2H4S5

Explanation:

Since the total mass is 5.000g

Mass of sulphur = 5.000-(0.6357+0.1070)

Mass of sulphur = 4.2573g

Using Empirical relation

C= 0.6357 H= 0.1070 S= 4.2573

Divide through by their molar mass to obtain the smallest ratio

C= 0.6357/12 H=0.1070/1 S=4.2573/32

C= 0.053 H= 0.1070 S= 0.133

Divide through by the smallest ratio (0.053)

C=0.053/0.053 H=0.1070/0.053 S=0.133/0.053

C=1 H=2 S=2.5

1:2:2.5 ,multiply through by 2 ,to obtain whole numbera

2:4:5

Therefore the empirical formula is C2H4S5. Thus only gives the ratio

Molecular formula is the chemical formula .

(Empirical formula) n = molecular formula

(C2H4S5)n = molar mass

[(12×2) + ( 1×4) +(32×5)]n = 188.4

188n=188.4

n= 1

Molecular formula = (C2H4S5)×1

Therefore the chemical formula of

lenthionine is C2H4S5

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

At the anode:

At oxidation reaction is:

The overall equation for the reaction is:

The overall cell potential is:

Using the formula for the Nernst equation:

where;

E = 0.66

(Zn^2+)=0.22 M

Then

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log  [H⁺]

4.118  = - 2  log  [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M =  0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 =  0.76 - 0.0592/2logQ

0.660 - 0.76  =  - 0.0592/2logQ

-0.1 =  - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 =  [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92  × 10^-3/[H^+]^2

[H^+]^2 = 7.92  × 10^-3/2.39 × 10^3

[H^+] = 1.82  × 10^-3

pH = -log[H^+]

pH = -log[ 1.82  × 10^-3]

pH = 2.7

Learn more: brainly.com/question/11897796