Answer:
horse latitudes
The horse latitudes are regions located at about 30 degrees north and south of the equator. These latitudes are characterized by calm winds and little precipitation. The horse latitudes are located at about 30 degrees north and south of the equator
Answer: In metallic bonds, the mobile electrons surrounding the positive ions are called <u><em>dipole</em></u>.
Answer:
P = 133.4 atm
Explanation:
Given data:
Mass of helium = 988 g
Volume = 45.0 L
Temperature = 23°C (23+273= 296 K)
Pressure of cylinder = ?
Solution:
Number of moles of helium:
Number of moles = mass/ molar mass
Number of moles = 988 g/ 4 g/mol
Number of moles = 247 mol
Pressure:
PV = nRT
P = nRT/V
P = 247 mol × 0.0821. atm. L/mol. K × 296 K / 45 L
P = 6002.5 / 45 L
P = 133.4 atm
A mixture of Cu2 and CuO of mass 8.828g is reduced to copper metal with hydrogen:
Cu2O + H2 --> 2Cu + H2O
CuO + H2 --> Cu + H2O
If the mass of pure copper isolated was 7.214g, determine the percent by mass of CuO in the original sample
Let x = grams of CuO in the original sample.
y = grams of Cu2O in the original sample.
Eq. #1 x + y = 8.828 grams
Molar mass of CuO = 63.5 + 16 = 79.5 grams
Moles of CuO = x ÷ 79.5
Molar mass of Cu2O = 63.546 + 32 = 95.5 grams
Moles of Cu2O = y ÷ 95.5
According to the 2nd balanced equation, CuO + H2 --> Cu + H2O ,
1 mole of CuO produces 1 mole of Cu.
So, x ÷ 79.5 moles of CuO will produce x ÷ 79.5 moles of Cu
According to the 1st balanced equation, Cu2O + H2 --> 2Cu + H2O,
1 mole of Cu2O produces 2 moles of Cu
So, (y ÷ 95.5) moles of Cu2O will produce 2 * (y ÷ 95.5) moles of Cu
Since, the mass of pure copper isolated was 7.214 grams
Moles of Cu = (7.214 ÷ 63.5)
Moles of Cu from Cu2O + moles of Cu from CuO = total moles of Cu!!
2 * (y ÷ 95.5) + (x ÷ 79.5) = (7.214 ÷ 63.5)
Multiply by both sides by 95.5 * 79.5 * 63.5 to get rid of denominators
(2 * 79.5 * 63.5) y + (95.5 * 63.5) x = (7.214 * 95.5 * 79.5)
10,096.5 y + 6,064.25 x = 36,418.0755
Divide both sides by 6,064.25
x + 1.665 y = 6
Eq.#2 x = 6 – 1.665 y
Eq. #1 x + y = 8.828
x = 8.828 – y
8.828 – y = 6 – 1.665 y
0.665 y = 2.828
y = 4.25 grams of Cu2O
x = 8.828 – 4.25 = 4.58 grams of CuO
% CuO = (4.58 ÷ 8.828) * 100 = 51.88% CuO