Answer: The molecular weight of the gas 28. The gas is probably N2
Explanation:Please see attachment for explanation
decameters - meters: multiply by 10
meters to meters: multiply by 1
centimeters to meters: divide by 100
millimeters to meters: divide by 1000
For the rows at the bottom:
hectometer row: 100, multiply by 100, 4500
decameter row: 10, multiply by 10, 450
meter row: 1, multiply by 1, 45
decimeter row: 0.1, divide by 10, 4.5
centimeter row: 0.01, divide by 100, 0.45
im guessing theres a millimeter row at the bottom:
millimeter row: 0.001, divide by 1000, 0.045
hope this helps!
Empirical formula: The formula consist of proportions of the elements which is present in the compound or the simplest whole number ratios of atoms.
Now, molecular formula is equal to the product of n (ratio) and empirical formula.
Molecular formula = 
(1)
molecular formula =
(given)
Since, 6 is the smallest subscript in above molecular formula to get the simpler whole number of atoms. Therefore, divide all the subscripts i.e. number of carbon atoms (12), number of hydrogen atoms (24) and number of oxygen atoms (6) by 6.
empirical formula becomes 
Thus, according to the formula (1)
Hence, empirical formula of given molecular formula is
<span>Answer:
1/4 is the average bond order for a pâ’o bond (such as the one shown in blue) in a phosphate ion.</span>
Here's the equation:
<span>Fe2 O3 + 2Al → 2Fe + Al2 O3
</span>
Here's the question.
What mass of Al will react with 150g of Fe2 O3?
<span>In every 2 moles Al you need 1 mole Fe2O3 </span>
<span>moles = mass / molar mass </span>
<span>moles Fe2O3 = 150 g / 159.69 g/mol </span>
<span>= 0.9393 moles </span>
<span>moles Al needed = 2 x moles Fe2O3 </span>
<span>= 2 x 0.9393 mol </span>
<span>= 1.879 moles Al needed </span>
<span>mass = molar mass x moles </span>
<span>mass Al = 26.98 g/mol x 1.879 mol </span>
<span>= 50.69 g </span>
<span>= 51 g (2 sig figs)
</span>
So the <span>mass of Al that will react with 150g of Fe2 O3 is 51 grams.</span>
