<h2>a)
The rate at which 
is formed is 0.066 M/s</h2><h2>b)
The rate at which molecular oxygen 
is reacting is 0.033 M/s</h2>
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of 
= ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
= 0.066 M/s
Rate in terms of disappearance of = ![-\frac{1d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D)
Rate in terms of appearance of = ![\frac{1d[NO_2]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7B2dt%7D)
1. The rate of formation of
![-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO_2%5D%7D%7B2dt%7D%3D%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
![\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B2%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.066M%2Fs)
2. The rate of disappearance of
![-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7B2dt%7D)
![-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
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The mass of the solute required is 250.25 g.
<h3>What is the mass of the solute?</h3>
We know that the number of moles of the solute can be used to obtain the mass of the solute that is required. We can now try to find the mass of the solute that is required.
Concentration of the solution = 0.350M
Volume of the solution = 6.5 L
Number of moles of the solute = 0.350M * 6.5 L
= 2.275 moles
We now have the mass of the solute as;
2.275 moles * 110 g/mol
= 250.25 g
Th measured mass of the solute that we would have to use is 250.25 g.
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Missing parts;
A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?
Questions:
The questions or computes to do are:
<span>a- a massa, em
kg, de cada placa de alumínio;
b- a quantidade mínima de viagens
necessárias para que apenas um veículo de transporte entregue o material
solicitado ao cliente.
Dado: densidade do alumínio = 2,7 g/cm3
Answer:
a) mass in kg of every aluminum plate
Dimensions of every aluminum plate: </span>
<span>2 M X 50 Cm X 2cm
Volume: 200 cm * 50 cm * 2 cm = 20,000 cm^3
Mass:
density = mass / volume => mass = density * volume = 2.7 g/cm^3 * 20,000 cm^3 = 54,000 g = 54 kg.
Answer: the mass of everyplate of aluminum is 54 kg.
b) number of travels required for one truck deliver all the material:
number of travels = amount requested / amount that a truck can deliver in one travel.
amount requested: 100 plates
mass of 100 plates = 100 plates * 54 kg / plate = 5,400 kg
limit of transport per travel: 3 tons = 3,000 kg
number of travels = 5,400 kg / 3,000 kg/travel = 1.8 travels => 2 travels.
Answer: at least 2 travels.
</span>
Newton's third law states that every action has an equal and opposite reaction. The action and reaction forces are pairs of opposing forces.
In the given examples all three obey Newton's third law.
B) Action force: John pulls the door handle
Reaction: door handle gets pulled
C) Action force: Tire pushes on road
Reaction: The road pushes on the tire, vehicle moves
D) Action force: Exhaust pushes out of a rocket
Reaction: Rocket is pushed forward
Ans A) All these are examples of Newtons third law
Answer:
C. Atoms of elements with five to seven valence electrons form anions in order to meet the octet rule.
Explanation:
- Atoms of elements gain or lose electron(s) to obey the octet rule by forming cations or anions.
- Atoms with 1 to 3 valence electrons lose electrons to form cations in order to attain a stable configuration.
- Atoms with 5 to 7 valence electrons gain electron(s) to form anions in order to attain stable configuration.
- However, atoms with 8 valence electrons do not require to gain or lose electrons since they an octet configuration.
- Atoms of metallic elements such as those in group 1 and 2 lose electron(s) to form cations while atoms of non-metallic elements such as halogens require to gain electron(s) to form anions so as to obey the octet rule.
