Answer:
32.7 kilograms of aluminium oxide will be produced.
Explanation:

Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )
Moles of aluminium = 
According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:
of aluminum oxide
Mass of 320.37 moles of aluminum oxides:
320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg
32.7 kilograms of aluminium oxide will be produced.
The relationship between pressure and solubility of the gas is given by Henry's law as:

where,
is the solubility of the gas.
is proportionality constant i.e. Henry's constant.
is pressure of the gas.
(given)
(given)
Substituting the values,

To convert
to
:
Molar mass of benzene,
= 

Now for converting into
:
Since, 
So,
.
Hence, the solubility of benzene in water at
in
is
.
To find the mass you need to find the weight of a mol of the molecules by adding up the atomic mass.
N = 14.007 g/mol
H = 1.008 g/mol
S = 32.065 g/mol
O = 16 g/mol
2(14.007) + 8(1.008) + 32.065 + 4(16) = 132.143 g/mol
Now you know how much an entire mol weight you multiply it by how much you actually have
0.00456 * 132.143 = 0.603 g
Answer: A
Explanation: The spindle fibers bring the chromosomes together in prophase then apart in telophase.
Answer:
5.5 atm
Explanation:
Step 1: Calculate the moles in 2.0 L of oxygen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
2.0 L × 1 mol/22.4 L = 0.089 mol
Step 2: Calculate the moles in 8.0 L of nitrogen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
8.0 L × 1 mol/22.4 L = 0.36 mol
Step 3: Calculate the total number of moles of the mixture
n = 0.089 mol + 0.36 mol = 0.45 mol
Step 4: Calculate the pressure exerted by the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 0.45 mol × (0.0821 atm.L/mol.K) × 298 K / 2.0 L = 5.5 atm