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QveST [7]
3 years ago
5

Label each of the components in the equation below what type of reaction

Chemistry
1 answer:
Citrus2011 [14]3 years ago
3 0

It is hard to answer this because not much information is given.

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What is the difference between a physical and chemical change?
Zanzabum
In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed. The distinction between physical and chemical change is not clear cut.
6 0
3 years ago
Read 2 more answers
How many molecules are in 0.500 mole of N2O5?
Oksi-84 [34.3K]

Answer:

3,011.10e23.

Explanation:

3 0
2 years ago
g Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane i
Kamila [148]

Answer:

6.05g

Explanation:

The reaction is given as;

Ethane + oxygen --> Carbon dioxide + water

2C2H6 + 7O2 --> 4CO2 + 6H2O

From the reaction above;

2 mol of ethane reacts with 7 mol of oxygen.

To proceed, we have to obtain the limiting reagent,

2,71g of ethane;

Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol

3.8g of oxygen;

Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol

If 0.0903 moles of ethane was used, it would require;

2 = 7

0.0903 = x

x = 0.31605 mol of oxygen needed

This means that oxygen is our limiting reagent.

From the reaction,

7 mol of oxygen yields 4 mol of carbon dioxide

0.2375 yields x?

7 = 4

0.2375 = x

x = 0.1357

Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g

8 0
3 years ago
Traumatic acid contains 63.13% carbon, 8.830% hydrogen, and 28.03% oxygen. its molar mass is 228 g/mol. determine the empirical
Zina [86]
Q1)
firstly we need to determine the empirical formula of the compound. empirical formula is the simplest ratio of components in the compound.
percentages of the elements have been given, so lets assume we are calculating for a compound of 100g 
                                  C                               H                              O
mass                       63.13 g                    8.830 g                       28.03 g
molar mass             12 g/mol                   1 g/mol                      16 g/mol
number of moles     63.13/12                   8.830/1                      28.03/16
                                5.26                           8.830                           1.75
divide by the smallest number of moles 
                                5.26/1.75                   8.830/1.75                   1.75/1.75
                                = 3.01                        = 5.04                           =1
rounded off to the nearest whole numbers 
C - 3
H - 5
O - 1
therefore empirical formula = C₃H₅O

Q2)
we have to next determine the molecular formula of the compound 
molecular formula gives the actual composition of elements in the compound. 
since we know the empirical formula and molecular mass, we can find how many empirical units are in the molecular formula.
mass of empirical unit = Cx3 + Hx5 + Ox1
                                    = 12 g/mol x 3 + 1g/mol x 5 + 16 g/mol x 1
                                    = 36 + 5 + 16 = 57 g/mol
the molecular mass = 228 g/mol
then number of empirical units in the molecular formula = 228 / 57 = 4
therefore there are 4 empirical units 
then the molecular formula = 4 x empirical formula =4 (C₃H₅O)
molecular formula = C₁₂H₂₀O₄




3 0
3 years ago
!!!Need answer for chem homework ASAP PLS !!!!
stealth61 [152]

Answer:

The answer to your question is: kc = 6.48

Explanation:

Data

             Given                 Molecular weight

CaO =    44.6 g                    56 g

CO₂ =     26 g                       44 g

CaCO₃ = 42.3 g                  100 g

Find moles

        CaO                 56 g ----------------  1 mol

                                 44.6 g --------------   x

                               x = (44.6 x 1) / 56 = 0.8 mol

        CO₂                 44 g -----------------  1 mol

                                26 g ----------------    x

                               x = (26 x 1 ) / 44 = 0.6  moles

        CaCO₃            100 g ---------------   1 mol

                                42.3g --------------    x

                              x = (42.3 x 1) / 100 = 0.423 moles

       

Concentrations

 

        CaO     =    0.8 / 6.5  = 0.12 M

         CO₂    =     0.6 / 6.5 = 0.09 M

        CaCO₃  =   0.423 / 6.5 = 0.07 M

Equilibrium constant =   \frac{[products]}{[reactants]}

Kc = [0.07] / [[0.12][0.09]

Kc = 0.07 / 0.0108

kc = 6.48

7 0
3 years ago
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