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loris [4]
3 years ago
7

A reaction expected to produce 245.6 g of a product produces 122.1 g of the

Chemistry
1 answer:
slava [35]3 years ago
4 0

Answer:

50%

Explanation:

% yields =(experimental yields / theoretical yields ) x100%

            =    122.1÷  245.6 ) x100% = 50%

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Which of the following rocks would probably be made up of the largest
IrinaVladis [17]

Answer:

Granite will be made up of the largest

8 0
3 years ago
If 2.68 g of benzaldehyde are involved with the mixed aldol condensation reaction, how many moles of benzaldehyde are present? R
Zielflug [23.3K]

Answer:

The number of moles of benzaldehyde = 0.0253 moles

Explanation:

The molecular formula of benzaldehyde is C₇H₆O

Its molecular mass is calculated from the atomic masses of the constituent atoms.

C = 12.0 g: H = 1.0 g; O = 16.0 g

Molecular mass = ( 12 * 7) + (1 * 6) + (16 * 1) = 106.0 g/mol

Number of moles of  substance = mass of substance/ molar mass of the substance

mass of benzaldehyde = 2.68; molar mass = 106.0 g/mol

Number of moles of benzaldehyde = 2.68 g/ 106 g/mol = 0.0253 moles

Therefore, the number of moles of benzaldehyde = 0.0253 moles

8 0
3 years ago
2.<br> An alkane has at least on C=C<br> bond.<br> ut of<br> Select one:<br> O True<br> O False
garik1379 [7]
I think it’s false?????
8 0
4 years ago
Consider the balanced chemical reaction below and determine the percent yield of sodium bromide if 2.36 moles of iron(iii) bromi
hammer [34]
        First   the  theoretical yield   of Nabr
  by use  of  mole  ratio   between  FeBr3  and  NaBr  which  is  2:6   the   theoretical yield  

  =2.36  x6/2= 7.08  moles

the  %  yield  =  actual  yield/  theoretical  yield  x  100

that  is    6.14/7.08  x100=  86.72%
8 0
4 years ago
Read 2 more answers
1) Given the balance equation below. Calculate how much Na3PO4 in grams you
juin [17]

Answer:

<u>136.67 g of Na3PO4 i</u>s required to create 100 gram of NaOH.

Explanation:

The balanced equation:

Na_{3}PO_{4} + 3 KOH \rightarrow 3 NaOH + K_{3}PO_{4}

1 mole Na3PO4 = 164 g/mole (Molar mass)

1 mole NaOH = 40 g/mole (Molar mass)

Now,

1 mole of Na3PO4 produce = 3 mole of NaOH

164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH

or

120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4

1 g/mol of NaOH is produced from =

\frac{164}{120}

100 grams of NaOH is produced from =

\frac{164}{120}\times100 gram of Na3PO4

calculate,

= 136.67 g

8 0
3 years ago
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