When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
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:
726 torr
Explanation:
Generally, atmospheric pressure can be measured using a manometer which is in form of a U-shaped tube. In addition, 1 mm Hg is equivalent to 1 torr. Therefore, 752 torr is equivalent to 752 mm Hg. Therefore, the total pressure will be equivalent to the atmospheric pressure (mm Hg) + the mercury height.
In this case, the mercury height = -26 mm
Thus:
The helium pressure = 752 - 26 = 726 mm Hg
This is also equivalent to 726 torr
Answer:
18.76atm
Explanation:
Using the formula V1P1/T1 = V2P2/T2, from combined gas law. Volume is constant since we have not been given. Therefore the formula comes to be; P1/T1 = P2/T1
To get P2 = T2(P1/T1)
Where P2 is final pressure
P2 = 239K ( 23atm/293K)
=18.76atm
Answer:
109° 27'
Explanation:
The ammonium ion is tetrahedral in shape, all the HNH bonds are exactly at the tetrahedral bond angle since there are only bond pairs in the structure and no lone pairs. Recall that lone pairs decrease the bond angke from the ideal value in a tetrahedron due to higher repulsion.
