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Ratling [72]
3 years ago
8

Draw the structure for each compound below:________. A) Propanedial B) 4-Phenylbutanal C) (S )-3-Phenylbutanal D) 3,3,5,5-Tetram

ethyl-4-heptanone E) (R )-3-Hydroxypentanal F) meta-Hydroxyacetophenone G) 2,4,6-Trinitrobenzaldehyde H) Tribromoacetaldehyde I) (3R,4R )-3,4-Dihydroxy-2-pentanone

Chemistry
1 answer:
djverab [1.8K]3 years ago
7 0

Answer:

Please check attachment

Explanation:

In this question, we are to draw the structure for each of the compounds below.

Please check attachment for the diagrams

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You are planting flowers in large pots. You plan to plant flowers in 10 pots. 15 cups of potting soil will fill one pot. A bag o
dimaraw [331]

Answer:

I will need  six (6) bags of potting soil

Explanation:

Since you plan on planting in 10 pots and need 15 cups of potting soil per pot, the total amount of potting soil you need <em>(in cups)</em> is 10 X 15 = 150 cups of potting soil.

We have that a bag contains 25 sups. to get the number of bags needed, we have to divide 150 by 25. This will give us 150 / 25 = 6 bags.

Therefore, I will need  six (6) bags of potting soil

3 0
4 years ago
I build a fire and burn 3 logs. Apply the Law of Conservation of Matter
photoshop1234 [79]
Capture all of the smoke and weight it. it will weigh exactly the same before and after you burn it but will just be CO2 and H2O gas.
7 0
3 years ago
A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
Paul [167]

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

5 0
4 years ago
Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be
serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

                  3 CuSO₄   +  2 Al   ⇒   3 Cu   +  Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

                160 g of CuSO₄  ----------------- 1 mol

                156 g                   -----------------  x

                      x = (156 x 1) / 160

                      x = 0.975 moles

               27 g of Al -------------------------- 1 mol

               29 g of Al -------------------------- x

                x = (29 x 1)/27

                x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical     3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental  0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.      

                    3 moles of CuSO₄ ------------------ 3 moles of Cu

                 0.975 moles of CuSO₄ ---------------  x

                         x = (0.975 x 3)/3

                        x = 0.975 moles of Cu were obtained.        

3 0
3 years ago
Read 2 more answers
What is the mass of 0.251 moles of water? Using dimensional analysis show work
AVprozaik [17]

1 \text{ mol of H}_2 \text{O} \equiv 18 \text{ g} \implies 0.251  \text{ moles of H}_2 \text{O} \equiv 4.518 \text{ g}

3 0
3 years ago
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