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The mass of a crucible and a hydrated salt was found to be 21.447 g. The mass of the crucible and the anhydrous salt was 20.070 g. The mass of the crucible was 17.985 g. How do you calculate the mass of the hydrate heated?
The question asked you to determine the mass of the hydrate, even though you could calculate the mass of anhydrous salt. And this is simply the intial mass of the...
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The atomic weight of aluminum is 26.982 u, how many aluminum atoms are there in a 4.55g sample of aluminum?
The key to this problem is the correct interpretation of what atomic weight means. Notice that aluminium is said to have an of $"26.982 u"$. The unit used here...
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A 50.0-kg mixture of salt and water is 15.0% (by mass) salt. What is the mass of salt and water in the mixture?
You know that your mixture is $15.0%$ salt by mass, so make sure that you understand what that means before moving forward. For a component of a mixture, its...
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How many nitrate ions, NO3-, and how many oxygen atoms are present in 1.00 microgram of magnesium nitrate, Mg (NO3)2? Also what mass of oxygen is present in 1.00 microgram of magnesium nitrate?
Number of nitrate ions $"Moles of Mg"("NO"_3)_2 = 1.00 × 10^"-6" color(red)(cancel(color(black)("g Mg"("NO"_3)_2))) × ("1 mol Mg"("NO"_3)_2)/(148.31 color(red)(cancel(color(black)("g Mg"("NO"_3)_2)))) = 6.743 × 10^"-9"color(white)(l) "mol Mg"("NO"_3)_2$ $"Moles of NO"_3^"-" = 6.743...
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What is (i) $%O$ in $"aluminum sulfate"$ by mass? And (ii) given a $1.00*g$ mass of this salt, how many formula units does this mass represent?
$O%=(12xx16.00*g*mol^-1)/(342.15*g*mol^-1)xx100%=??%$ For formula units present in $1.000*g$ of the salt, we work out the molar quantity....... $"Moles of aluminum sulfate"=(1.000*g)/(342.15*g*mol^-1)=2.923xx10^-3*mol$ And since we know that $1*mol$ contains $N_A="Avogadro's Number"=6.022xx10^23*"particles"*mol^-1$, we...
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Aluminum oxide is decomposed to extract pure aluminum from the mineral. If 6.7 tons of aluminum oxide is mined, calculate the mass of pure aluminum produced?
1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms...
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A solution is prepared from $25*g$ of salt dissolved in $1*L$ of water. If the salt is an hydrate of sodium thiosulfate, how many water molecules are associated with this salt?
$Na_2S_2O_3*5H_2O$ is the common laboratory reagent, this so-called pentahydrate forms large sugary crystals so it is easy and convenient to handle. The formula mass of this material is $248.18*g*mol^-1$ (i.e....
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Given a $1*L$ volume of solution that is $1.25*mol*L^-1$ with respect to $MgCl_2$, there are...?
$"a. 2 moles of ions in solution."$
$"b. 1.25 moles of ions in solution."$
$"c. 2.5 moles of ions in solution."$
$"d. 2 moles of ions in solution."$
The $"mole"$, $N_A$, is simply a number. Admittedly, it is a very large number, $N_A=6.022xx10^23*mol^-1$. And if you have a mole of stuff, you have $6.022xx10^23$ INDIVIDUAL ITEMS of that...
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A chemist has 400 grams of salt solution that is 10% salt. How many grams of 20% salt solution must be added to obtain a 12% solution of salt?
So, you're dealing with two salt of different . Start by calculating how much salt you get in the 400-g sample of the 10% solution. $m_"salt"/m_"solution" * 100...
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Are $"stannic ions"$ $Sn^(4+)$ or $"plumbic ions"$ $Pb^(4+)$ oxidizing or reducing agents? What about $"stannous ions"$ $Sn^(2+)$ or $"plumbous ions"$ $Pb^(2+)$?
The metal in the higher oxidation state is the reducing agent.......the electron pair acceptor: And of course, these species could be reduced to give the bivalent metal ion.......... $SnX_4 +2e^(-)...
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