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If you put an object that weighs 8 grams on one side of a balance scale, you would have to put about 20 paper clips on the other side to balance the weight. How many paper clips would balance the weight of a 10-gram object?
You're basically saying that 8 grams is equivalent to 20 paper clips. Therefore 10 grams would be equivalent to (10/8) x 20 = 25 paper clips. Or alternatively you could...
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Which would have more density - a piece of wood with a mass of 250 grams or 1,500 grams of the SAME TYPE of wood?
By definition.... $rho,$ $"density"$ $=$ $"mass"/"volume"$, and thus is an INTENSIVE material property rather than an extensive property such as mass or charge. Is the density likely to be...
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Potassium is a very flammable metal if it comes in to contact with water. As it burns with the water it creates potassium hydroxide ($KOH$) If you separate the potassium out from 50 grams of $KOH$, how many grams of potassium would you have?
$"Moles of potassium hydroxide"$ $=$ $(50*g)/(56.11*g*mol^-1)$ $"Mass of potassium metal"$ $=$ $(50*g)/(56.11*g*mol^-1)xx39.10*g*mol^-1$ $~=$ $??g$
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In the reaction: $A + B -> C$, if 4.2 grams of A react to make 6.8 grams of C, then how many grams of B were reacted?
We can use the Law of Conservation of Mass to solve the problem. $underbrace("A")_color(red)("4.2 g") + underbrace("B")_color(red)(x) → underbrace("C")_color(red)("6.8 g")$ $"4.2 g"color(white)(l) + x = "6.8 g"$ $x = "6.8...
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100 grams of magnesium and 2.50 grams of oxygen react together in synthesis reaction to form magnesium oxide. How do you write this balanced reaction?
It is abundantly clear that magnesium is in vast molar excess: i.e. $"Moles of metal"$ $=$ $(100*g)/(24.31*g*mol^-1)$ $~=$ $5*mol$ $"Moles of dioxygen"$ $=$ $(2.5*g)/(32.00*g*mol^-1)$ $~=$ $0.08*mol$ Clearly, $O_2$ is the...
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How many grams of carbon monoxide can you produce from 3 grams of carbon and an excess of oxygen?
We need (i) a stoichiometric equation: $C(s) +1/2O_2(g) rarr CO(g)$ And now (ii) we need the equivalent quantity of $C$. $"Moles of carbon"$ $=$ $(3.0*g)/(12.011*g*mol^-1)$ $=$ $1/4*"mole"$. So mass of...
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In the combustion of 5.00 grams of an alcohol the chemist produced 11.89 grams of $"CO"_2$ and 6.09 grams of $"H"_2"O"$. What is the identity of the alcohol?
Let's first write the chemical equation for this combustion reaction: $"alcohol"color(white)(l) + "O"_2(g) rarr "CO"_2(g) + "H"_2"O"(g)$ What we can do first is use the molar masses of...
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Hydrogen and oxygen react chemically to form water. How much water would form if 4.8 grams of hydrogen reacted with 38.4 grams of oxygen?
We need a stoichiometric equation for water synthesis: $H_2(g) + 1/2O_2(g) rarr H_2O(l)$ Clearly, dihydrogen must be present in a 2:1 molar ratio with respect to dioxygen. $"Moles of dihydrogen"$...
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If 5.563 grams of $CH_4$ reacted with 25.00 grams of oxygen, how much $CO_2$ , in grams, can be produced? $CH_4(g) + 2O2(g) -> CO_2(g) + 2H_2O(g)$?
$"Moles of methane"$ $=$ $(5.563*g)/(16.043*g*mol^-1)$ $=$ $0.3468*mol$. $"Moles of dioxygen"$ $=$ $(25.00*g)/(31.998*g*mol^-1)$ $=$ $0.7813*mol$. Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation: $CH_4(g) +...
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$C_2H_4 + 3O_2 -> 2CO_2 +2H_2O$. If you start with 45 grams of ethylene ($C_2H_4$), how many grams of carbon dioxide will be produced?
$"No. of mol of ethylene"=45/(2*12+4*1)~~1.607mol$ $"No. of mol of "CO_2" produced"=2*1.607=3.214mol$ $"Mass of "CO_2"=3.214*(12+2*16)=141.416g$
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