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What determines if in a reaction between sodium chloride and sulfuric acid you will get sodium bisulfate or sodium sulfate?
If 2 equiv sodium hydroxide are used: $H_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2H_2O(l)$ If 1 equiv sodium hydroxide is used: $H_2SO_4(aq) + NaOH(aq) rarr NaHSO_4(aq) + H_2O(l)$ Both reactions...
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Sodium chloride reacts with copper sulfate to produce sodium sulfate and copper chloride. $2NaCI (aq) + CuSO_4 (aq) -> Na_2SO_4 (aq) + CuCl_2 (s)$. What type of reaction is this?
Aqueous sodium chloride, $"NaCl"$, will not react with aqueous copper(II) sulfate, $"CuSO"_4$, because the two potential products are soluble in aqueous solution. The chemical equation given to you is...
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Which option contains a $2*g$ mass of sodium atoms....
$1.$ $5.24xx10^22*"sodium atoms?"$
$2.$ $6.022xx10^24*"sodium atoms?"$
$3.$ $3.12xx10^23*"sodium atoms?"$
$4.$ $5.24xx10^22*"sodium atoms?"$
We know that $1*mol$ contains $6.022xx10^23$ individual sodium atoms, and has a mass of $22.99*g$. And thus for $1.$ there is a $2*g$ mass of $Na$. For $2.$ there is...
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How many grams of sodium sulfide are formed if 1.50 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 94.0 % yield?
The first step for any chemical reaction is to write out the chemical equation to determine the molar ratios of the reactants and products. Then we will use the molecular...
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A $16.7*g$ mass of sodium sulfate is dissolved in a volume of $0.125*L$ of water. What is the concentration with respect to sodium sulfate?
$"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$ $=$ $(16.7*g)/(142.04*g*mol^-1)xx1/(0.125*L)$ $=$ $0.940*mol*L^-1$ with respect to sodium sulfate. Note that the way we have set out the problem, appropriate units are...
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A $5*g$ mass of sodium chloride, and an $8.2*g$ mass of sodium sulfate are dissolved in a $200*mL$ of aqueous solution. What is $[Na^+]$?
$"Moles of NaCl"=(5.0*g)/(58.44*g*mol^-1)=0.0856*mol$. $"Moles of"$ $Na_2SO_4=(8.2*g)/(142.04*g*mol^-1)=0.0577*mol$. And thus $[Na^+]=(0.0856*mol+2xx0.0577*mol)/(200*mLxx10^-3*L*mL^-1)$ $=??*mol*L^-1$. Why did I double the molar quantity with respect to $Na_2SO_4$. Did I make a mistake?
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How many liters of a solution containing 45% acid must be mixed with a solution containing 60% acid to obtain 40 liters of a 48% acid solution?
In order to solve this problem, you need to set up two equations with two unknowns, the volume of the and the volume of the solutes. Use the...
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A chemist has one solution containing 30% insecticide and another solution containing 50% insecticide. How much of each solution should the chemist mix to get 200 L of a 42% insecticide?
Amount of substance = concentration x volume. Let $x$ = the volume of the $30%$ solution. Let $y$ = the volume of the $50%$ solution. Because the total amount of...
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A solution is made containing 11.2g of sodium sulfate per 95.0g of water. What is the percent by mass of the solute?
$"Solution percentage by mass"$ $=$ $"Mass of solute"/"Mass of solution"$ $xx100%$ $=$ $(11.2*g)/(11.2*g+95.0*g)xx100%$ $~=10%$
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Ksp=2.4x10^-5 for calcium sulfate.
What is the molar solubility of calcium sulfate in pure water?
What is the mass solubility of calcium sulfate in pure water,
expressed in g/L?
(a) Calcium sulfate is sparingly soluble. In a saturated solution the aqueous ions are in equilibrium with the insoluble salt: $sf(CaSO_(4(s))rightleftharpoonsCa_((aq))^(2+)+SO_(4(aq))^(2-))$ For which $sf(K_(sp)=[Ca_((aq))^(2+)][SO_(4(aq))^(2-)]=2.4xx10^(-5)color(white)(x)"mol"^2."l"^-2)$ The solubilty $sf(s)$ of the salt...
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