.help!!! due today!!! what is happening and how it is demonstrating Newton’s 1st law. make sure you describe inertia and how it
relates to your demonstration.
okayyy so ive got to make a video and im planning on making this but then i have to explain how is it demonstrating Newton's 1st law and then I have to describe how Inertia relates to this demonstration, please help a girl out!!!
okayyy so ive got to make a video and im planning on making this but then i have to explain how is it demonstrating Newton's 1st law and then I have to describe how Inertia relates to this demonstration, please help a girl out!!!
1 answer:
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50 grams or 50,000 mili grams is the mass of solute in 1000 grams of a solution having a concentration of 5 parts per million.
Explanation:
Total mass of solution = 1000 grams or 1000 ml since 1 gram = 1 ml
concentration is 5 parts per million ( 5 mg in 1000 ml solution or 0.005 gram in 1000 ml)
the formula used for parts per million:
parts per million =
putting the values in the equation:
parts per million =
0.005 x 1000 = mass of solute
50 grams= mass of solute
converting this into mg
50,000 mg. is the total mass of solute in 5ppm of 1000 ml solution.
Answer:
Kp = 0.022
Explanation:
<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>
<em />
The equilibrium of ammonia occurs as follows:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where Kp is defined as:
<em>Where P represents partial pressure of each gas.</em>
<em />
As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:
P(NH₃) = 2.3atm - 2X
<em>Where X represents reaction coordinate</em>
<em />
Thus, pressure of hydrogen and nitrogen is:
P(N₂) = X
P(H₂) = 3X.
As partial pressure of hydrogen is 0.69atm:
3X = 0.69
X = 0.23atm:
P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm
P(N₂) = 0.23atm
P(H₂) = 0.69atm
Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.
We have the following equation representing the half-life decay:
A is the resulting amount after t time
Ao is the initial amount = 50 mg
t= Elapsed time
t half is the half-life of the substance = 14.3 days
We replace the know values into the equation to have an exponential decay function for a 50mg sample
That would be the answer for a)
To know the P-32 remaining after 84 days we have to replace this value in the equation:
So, after 84 days the P-32 remaining will be 0.85 mg