Answer:
The length of the wire = 352.66 feet.
Explanation:
A copper refinery produces a copper ingot weighing 150 lb. If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density of copper is 8.94 g/cm3. (Assume that the wire is a cylinder whose volume is V = πr2h, where r is the radius and h is its height or length.)
Step 1: Convert lb to kg
150 lb = 68.0389 kg
Step 2: Calculate volume of copper
Volume = mass / density
Volume = 68038.9 grams / 8.94 g/cm³
Volume = 7610.6 cm³ Cu
Step 3: Calculate length of wire
The diameter of the wire is 9.50 mm, so the radius is half of that (4.75 mm), or 0.475 cm.
The total "volume" of the wire is πr²h = (π)*(0.475 cm)²(h) = 0.708h = 7610 cm^3
7610 = 0.708h
h = 10749 cm = length of wire
The length of the wire = 352.66 feet.
Answer:
Part A:
"360 grams of NaCl can be dissolved in 1 L water. So, 2000 grams sugar can be dissolved in 1 L water then we can say that the solubility of salt is lesser in water as to sugar and both heightened by increasing the temperature. If we make a batch of 800 L we can add sugar, 1600 kg at 25 0c. We can add salt is 288 kg at 25 0c and the ingredient tomato is having low solubility."
Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/8061-describe-the-sequence-of-adding-ingredients-to-make-the-recipe.html
Part B:
'Manufacturers can generate new value minimize cost and increase operational stability by focusing on 4 broad areas; Management, Supply Circle, Product Design, and Value Recovery.'
Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/2807911-what-changes-could-be-made-to-optimize-the-manufacturing-process.html
Answer: It would be Proteins. So B
Hope this helps!!!
DNA makes RNA which makes Proteins
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Molar mass of NH_3
We know.
No of moles=Given mass/Molar mass
Now
Lets write the balanced equation
- There is 2moles of Ammonia
- 3moles of H_2
- 1mole of N_2
Now
For Hydrogen
For Ammonia
For Nitrogen
