Al₂O has a higher percentage of aluminium.
Explanation:
To solve this problem, we have to compare the molar mass of the aluminium in each compound to one another as percentage of the whole compound:
Molar mass of Al₂O = 2(27) + 16 = 70g/mol
Molar mass of Al(NO₃)₃ = 27 + 3[14 + 3(16)] = 213g/mol
Percentage by mass of Al in Al₂O =
x 100 = 77%
Percentage by mass of Al in Al(NO₃)₃ =
x 100 = 12.7%
Al₂O has a higher percentage of aluminium.
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Here we go ~
1 mole of
has 6.022 × 10²³ molecules of the given compound.
So, 0.78 mole of
will have ~


Answer:
B. halocline
Explanation:
it is a zone in the oceanic water that changes depending on the depth
Hope This Helped Sorry If Wrong
Answer:
625.46 °C
Explanation:
We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 19 °C
T(K) = 19 °C + 273
T(K) = 292 K
Next, we shall determine the Final temperature. This can be obtained as follow:
Initial volume (V₁) = 3.25 L
Initial temperature (T₁) = 292 K
Final volume (V₂) = 10 L
Final temperature (T₂) =?
V₁/T₁ = V₂/T₂
3.25 / 292 = 10 / T₂
Cross multiply
3.25 × T₂ = 292 × 10
3.25 × T₂ = 2920
Divide both side by 3.25
T₂ = 2920 / 3.25
T₂ = 898.46 K
Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 898.46 K
T(°C) = 898.46 – 273
T(°C) = 625.46 °C
Therefore the final temperature of the gas is 625.46 °C
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