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bagirrra123 [75]
3 years ago
9

For the reaction below:this question has multiple parts. work all the parts to get the most points.draw the major organic produc

t(s) in the sketch pad below. will the product mixture rotate the plane of polarization of plane-polarized light?
Chemistry
1 answer:
Allushta [10]3 years ago
3 0
Where is the reaction? The whole question didn't make any sense! Next time please put some time on what you want to ask.
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The element oxygen, represented by the symbol O, is classified as
Katarina [22]
Oxygen<span> is a </span>chemical element<span> having a symbol </span>O<span> and having an </span>atomic number<span> 8.
Oxygen is a pure substance , and it is not a mixture or solution, Niether it is a compound ... Instead , it forms chemical compounds by reacting with other elements.
So , according to above explanation ,
A. A pure substance , is the correct answer.</span>
3 0
4 years ago
Read 2 more answers
C + O2 → CO2 ,
DerKrebs [107]
First, if you want to get the amount of production from the equation, you need to use the units of mol. So, we need to know the molar mass of C which is 15.75/12=1.31 mol. Then we need to use the less part(O2) to get the production amount which is 0.116 mol. Then we can get the answer in units of g: 0.116*(12+16*2)=5.104 g.
5 0
4 years ago
Unit 1 chemistry test
Artist 52 [7]
Ok what’s the question sir
5 0
3 years ago
If 18.0 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentra
k0ka [10]

<em>Answer:</em>

  • The molarity of ammonia will be 2.88 M.

<em>Chemical equation</em>

                                      HCl +  NH3 ------> NH4Cl

First of calculate the moles of HCl

mole of HCl = Molarity × Vol (L)

mole of HCl = 0.800× 0.018 = 0.014 mole

As the in balance chemical, moles of HCl and NH3 areequal

so

                        moles of NH3= 0.014

                       Molarity of NH3 = moles ÷ V(L) = 0.014/0.005 =  2.88 M

<em>Result</em>:

  • The molarity of ammonia will be 2.88 M.
3 0
3 years ago
How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2S
Rama09 [41]

Answer: 2.796 grams

Explanation:

Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles

\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles

According to stoichiometry:

1 mole of BaCl_2 reacts with 1 mole of Na_2SO_4

0.012 moles of BaCl_2 will react with=\frac{1}{1}\times 0.012=0.012moles of Na_2SO_4

Thus BaCl_2 is the limiting reagent as it limits the formation of product. Na_2SO_4  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of BaCl_2 produces 1 mole of BaSO_4

0.012 moles of BaCl_2 will produce=\frac{1}{1}\times 0.012=0.012moles of BaSO_4

Mass of BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g

Thus 2.796 grams of BaSO_4  are produced.

6 0
4 years ago
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