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astra-53 [7]
3 years ago
5

2 C2H6 + 7 026 H2O + 4 CO2

Chemistry
1 answer:
Kay [80]3 years ago
3 0

Answer:

Oxygen is limiting reactant

Explanation:

Based on the chemical reaction:

2C2H6 + 7O2 → 6H2O + 4CO2

<em>2 mole of ethane reacts with 7 moles of oxygen</em>

<em />

For a complete reaction of 5.25 moles of ethane are required:

5.25 moles Ethane * (7mol Oxygen / 2mol Ethane) = 18.38 moles of oxygen

As there are just 15.0 moles of oxygen

<h3>Oxygen is limiting reactant</h3>
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Fill in the blanks with the correct term.
zzz [600]

Answer:

Fill in the blanks with the correct term.

a. a liquid that dissolves another substance.

b. a chemical that is dissolved.

c. a value used to describe the amount of one substance dissolved in another.

d. a liquid consisting of one substance dissolved in another.

Explanation:

a. A liquid that dissolves another substance is called the solvent.

b. A chemical that is dissolved solute.

c. A value used to describe the amount of one substance dissolved in another is called concentration.

d. A liquid consisting of one substance dissolved in another is called a solution.

4 0
3 years ago
The layer in which all living things on earth live and allows you to breathe is called the?
Ksju [112]
A troposphere because that is where all the CO2 and where we live
6 0
3 years ago
Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. The o
Nady [450]

Answer:

0.825 M

Explanation:

The osmotic pressure is a colligative property, that can be calculated using the following expression.

π = M × R × T

where,

π is the osmotic pressure

M is the molarity

R is the ideal gas constant

T is the absolute temperature (24°C + 273 = 297 K)

M = π / R × T = 20.1 atm / (0.08206 atm.L/mol.K) × 297 K = 0.825 M

5 0
3 years ago
nswer the following questions relating to HCl, CH3Cl, and CH3Br.HCl(g), can be prepared by the reaction of concentrated H2SO4(ag
alexdok [17]

Answer:

It is an example of double displacement reaction.

4.8 g of NaCl is needed to react.

Explanation:

Balanced reaction: H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)

Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.

In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.

As H_{2}SO_{4}(aq.) remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.

Molar mass of HCl = 36.46 g/mol

So, 3.0 g of HCl = \frac{3.0}{36.46} mol of HCl = 0.082 mol of HCl

According to balanced equation-

2 moles of HCl are produced from 2 moles of NaCl

So, 0.082 moles of HCl are produced from 0.082 moles of NaCl

Molar mass of NaCl = 58.44 g/mol

So, mass of 0.082 moles of NaCl = (0.082\times 58.44) g = 4.8 g

Hence 4.8 g of NaCl is needed to react.

4 0
3 years ago
In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

4 0
3 years ago
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