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:
Explanation:
Method 1 proportion
1 mole of chromium is 52 grams
11.9 moles = x grams
1/11.9 = 52/x Cross multiply
x = 11.9 * 52
x = 618.8 grams
Now I have used an approximate mass for Chromium. The answer you get here is expected to reflect the weigth given on your periodic table Use that to get your answer. You should give a number very close to mine. Round to 3 places as in 619.
Method Two Formula
mols = given mass / molecular mass
11.9 = given mass / 51.9961 Multiply both sides by 51.9961
11.9 *51.9961 = given mass
given mass = 618.75
given mass = 619
Answer:
NH4Br + AgNO3 —> AgBr + NH4NO3
Explanation:
When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:
NH4Br(aq) + AgNO3(aq) —>
In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:
NH4Br(aq) —> NH4+(aq) + Br-(aq)
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
The double displacement reaction will occur as follow:
NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)
NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)
This is what I found!!! I hope this helps!!!!
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or 
)
Given : kf = 1.86 
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907 )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
