Answer:
24.525 g of sulfuric acid.
Explanation:
Hello,
Normality (units of eq/L) is defined as:
Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:
Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:
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The question is incomplete, the complete question is;
AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —
aluminum bromide
monoaluminum tribromide
aluminide bromine
aluminum tribromide
Answer:
aluminum bromide
Explanation:
Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.
We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.
Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.
C Linnaeus was the first person known to have used the terms genus and species when classifying organisms.
1 2 3 4 5 and how the question ask
Answer:
Explanation:
Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :)
