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bixtya [17]
3 years ago
13

"a reaction was performed in which 0.57 g of 2-naphthol was reacted with a slight excess of 1-bromobutane to make 0.32 g of 2-bu

toxynaphthalene. calculate the theoretical yield and percent yield for this reaction."
Chemistry
1 answer:
murzikaleks [220]3 years ago
8 0
Convert 0.47g 2-naphthol to mols. mol = g/molar mass. 
Using the coefficients in the balanced equation, convert mols 2-naphthol to mols 2-butoxynaphthalene then you have to Convert mol 2-butoxynaphthalene to grams. g = mols x molar mass. This is the theoretical yield. 
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How much 6.0M HNO(small 3) is needed to neutralize 39mL of 2 M KOH
Digiron [165]
Equation of reaction 
<span>KOH+HNO3--->KNO3+H2O </span>
<span>CA=6.0M, CB=2.0M, VA=?, VB=39ml, na=1, nb=1 </span>
<span>CAVA/CBVB=na/nb </span>
<span>6*VA/2*39=1/1 </span>
<span>6VA/78=1 </span>
<span>6VA=78 </span>
<span>VA=78/6. VA=13ml. </span>
<span>VA=13ml.</span><span>
Hope this helped!!!!! </span>Don't forget Brainliest!!!!! You can always PM me if you don't want to waste any points!!!!!
8 0
3 years ago
Empirical Formula of P3O4H2?
puteri [66]

Answer:

H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.

Explanation:

3 0
3 years ago
Which question can be answered using the scientific method ?
anyanavicka [17]
The correct answer is D because the the rest will have different answers depending on the person you ask. D can be proven with facts by research
8 0
3 years ago
A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter
Ira Lisetskai [31]

Answer:

m H2O = 56 g

Explanation:

  • Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

5 0
3 years ago
An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t
erik [133]

Answer:

\% Fe^{+2}=70%

Explanation:

Hello,

In this case, we could considering this as a redox titration:

Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}

Thus, the balance turns out (by adding both hydrogen ions and water):

Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-

Thus, by stoichiometry, the grams of Fe+2 ions result:

m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}

Finally, the mass percent is:

\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\  \% Fe^{+2}=70%

Best regards.

8 0
3 years ago
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