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Semenov [28]
3 years ago
10

Anyone really good with Organic chemistry

Chemistry
1 answer:
dlinn [17]3 years ago
3 0

Answer:

umm maybe give me you're question I'll try

You might be interested in
How can balancing equations support the law of conservation
amm1812

Answer:

detail is given below.

Explanation:

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

For example:

In given photosynthesis reaction:

6CO₂ + 6H₂O  → C₆H₁₂O₆ + 6O₂

The given equation is balanced chemical equation of photosynthesis. There are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

If equation is not balanced,

CO₂ + H₂O  → C₆H₁₂O₆ + O₂

It can not follow the law of conservation of mass because mass is not equal on both side of equation.

4 0
3 years ago
How many atoms of germanium<br> are present in a sample<br> containing 1.65 moles Ge?
otez555 [7]

Answer:

9.9363 x 10^23  atoms

Explanation:

1.65   x   Avagadro's Number

1.65 * 6.022 x 10^23 =

5 0
2 years ago
Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in oxidation reduction reactions
borishaifa [10]

Answer: Oxidation is defined as the reaction in which there is loss of electrons. It is accompanied by increase in oxidation number.

Zn\rightarrow Zn^{2+}+2e^{-}

Zinc in solid state has oxidation number of zero and on losing electrons changes to Zn^{2+} with oxidation number of +2.

Reduction is defined as the reaction in which there is gain of electrons. It  is accompanied by decrease in oxidation number.

2H^++e^{-}\rightarrow H_2

Hydrogen ions with +1 oxidation state gains electrons and converts to molecular hydrogen with oxidation state of zero.

In the given redox reaction, both oxidation and reduction takes place. They go hand in hand.

Zn+H_2SO_4\rightarrow ZnSO_4 +H_2

Zinc converts to Zn^{2+} and gets oxidized. H^+ gains electron and convert to H_2.




4 0
3 years ago
A 25-mL solution of H2SO4 is completely neutralized by 18 mL of 1.0M NaOH. What is the concentration of the H2SO4?H2SO4(aq) +2Na
cricket20 [7]
1) Balanced chemical reaction

     H2SO4 (aq) + 2NaOH(aq) ----> Na2SO4(aq) + 2 H2O(liq)

2) Molar ratios

1 mol H2SO4 : 2 mol NaOH : 1 mol Na2SO4

3) Number of moles of NaOH


M = n / V => n = M * V = 1.0 M * 0.018 liter = 0.018 mol
of NaOH.


4) Determine the number of moles of H2SO4 using proportionality

1 mol H2SO4 / 2 mol NaOH = x / 0.018 mol NaOH

Solve for x:

x = 0.018 mol NaOH * 1 mol H2SO4 / 2 mol NaOH =0.009 mol H2SO4


5) Calculate the molarity using 0.009 mol and 25 mililiters

M = n / V = 0.009 mol / 0.025 liter = 0.36 M 

Answer: the concentration of the H2SO4 is 0.36 M



 


4 0
3 years ago
A 0.853-g of 90.0% lithium chlorate mixture decomposes with heat to give 313 ml of oxygen gas collected over water at 20 c and 7
SpyIntel [72]

Answer:

\%LiClO_3=90.4\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

LiClO_4\rightarrow LiCl+\frac{3}{2} O_2

It is widely known that when a gas given off from a reaction is collected over water, we can compute its pressure by minusing the total pressure 762 mmHg and the vapor pressure of water at the experiment's temperature (20 °C) in this case 17.5 mmHg as shown below:

p_{O_2}=762mmHg-17.5mmHg=744.5mmHg*\frac{1atm}{760mmHg}=0.980atm

Next, by using the ideal gas equation we compute the yielded moles of oxygen considering the collected 313 mL (0.313 L):

n_{O_2}=\frac{PV}{RT}=\frac{0.980atm*0.313L}{0.082\frac{atm*L}{mol*K}*293.15K}\\\\  n_{O_2}=0.0128molO_2

Now, via the 3/2:1 mole ratio between oxygen and lithium chlorate (molar mass = 90.39 g/mol), we compute the original mass of decomposed lithium chlorate as follows:

m_{LiClO_3}=0.0128molO_2*\frac{1molLiClO_3}{3/2molO_2}*\frac{90.39gLiClO_3}{1molLiClO_3}  \\\\m_{LiClO_3}=0.771gLiClO_3

Now, the percentage is computed as shown below:

\%LiClO_3=\frac{0.771g}{0.853g} *100\%\\\\\%LiClO_3=90.4\%

Best regards!

3 0
3 years ago
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