1. The benefits of been assertive include the following:
a. It improve one self image: When you choose to be assertive, you adopt a self realistic image.
b. It enhance how we understand others: Been assertive helps you to see others in a more realistic context.
c. Promote self awareness and self confidence: Been assertive helps you to develop a greater respect for your own points of view when dealing with issues.
d. Been assertive is a less stressful way of communicating.
2. The three steps needed to develop assertive communication, include:
a. Learn to 'no' more often: refuse to please everyone and do not behave according to people expectations of you. Be yourself.
b. Your tone must be properly turned when talking: Do not raise your voice or rush a communication. Be calm and cool when talking.
c. Be open to communication: Be willing to discuss issues until a suitable solution is found.
d. Pay attention to non-verbal communication of others: Keep eye contact when communicating and pay attention to other body gestures.
Answer:
A. respiration.
Explanation:
Cellular respiration can be defined as a series of metabolic reactions that typically occur in cells so as to produce energy in the form of adenosine triphosphate (ATP). During cellular respiration, high energy intermediates are created that can then be oxidized to make adenosine triphosphate (ATP). Therefore, the intermediary products are produced at the glycolysis and citric acid cycle stage.
Additionally, mitochondria provides all the energy required in the cell by transforming energy forms through series of chemical reactions; breaking down of glucose into Adenosine Triphosphate (ATP) used for providing energy for cellular activities in the body of living organisms.
Basically, oxygen goes into the body of a living organism such as plants, humans and animals when they breathe while glucose is absorbed by the body when they eat.
Hence, the conversion of sugar to energy in the presence of oxygen is respiration.
Via half-life equation we have:
Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:
There is 6.25 grams left of Ra-229 after 12 minutes.
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
