<span>26.833 liters
Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).
Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.
Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.
Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815
Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,
1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
</span>
<span><span>Atomic number36,</span><span>Atomic mass<span>83.80 g.mol -1,</span></span><span>Density<span>3.73 10-3 g.cm-3 at 20°C,</span></span><span>Melting point- 157 °C,</span><span>Boiling point<span>- 153° C</span></span></span>
No' of molecules divide by avogadro number , 6×6.023×10^23 so (2.2×10^22)÷(6.023×10^23)
= 0.03653 moles
moles × Molar mass = mass
n×Mr=m
0.03653×40 = 1.46 grams
The empirical formula is N₂O₅.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio¹ </u> <u> ×2² </u> <u>Integers</u>³
N 1.85 1 2 2
O 4.63 2.503 5.005 5
¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).
²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.
³Round off the number in the ratio to integers (2 and 5).
The empirical formula is N₂O₅.
Answer:
400.197mmHg
Explanation:
P1V1 / T1 = P2V2 / T2
Where P1=524 mm Hg V1 =275 ml T1 = 35°C +273 = 308k
V2= 325-ml T2= 5°C+273 = 278k , P2= ?
Substituting the values into the formula.
524 mm Hg ×275 ml /308k = P2×325-ml/278k
Cross multiply
524 mm Hg ×275 ml×278k=308k×P2×325-ml
40059800= 100100×P2
P2 = 40059800/100100
P2= 400.197mmHg
Hence, the second pressure will be 400.197mmHg
