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amid [387]
3 years ago
5

A chemist combines ethane and chlorine with the goal of producing chloroethane which other product could be reasonably expected

to form
Chemistry
2 answers:
lesya692 [45]3 years ago
5 0
<span>So when the chemist combines Ethane (CH3CH3) and Chlorine (Cl2) with the intention of producing Chloroethane (CH3CH2Cl), the other product that's formed in this reaction is 1,2-dichloroethane (ClCH2CH2Cl) also called as Ethylene dichloride with molecular weight of 98.954 g/mol. This is a colorless oily flammable substance that weighs heaver when vaporized.</span>
morpeh [17]3 years ago
3 0

Answer: other products are 1,2-dichloroethane, 1,1-dichloroethane and 1,1,1-trichloro ethane formed.

Explanation: since in chlorination of Alkane is an example for free-radical Substitution and due to more reactivity of free-radical excess chlorination (poly chlorination) Occurs.

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Explain why the product at the negative electrode is not always a metal.
Elena-2011 [213]

Answer:

Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series: the metal will be produced if it is less reactive than hydrogen. hydrogen will be produced if the metal is more reactive than hydrogen.

3 0
3 years ago
If you mix water and vinegar together, will it form a solid?
natka813 [3]
No, but. It will seperate into two different layers based on density
4 0
3 years ago
A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
While you are using a battery, the cell reaction is going a) forward, b) backward, c) at equilibrium, d) TEXASIN all possible
polet [3.4K]

Answer:Cell reaction is going forward.

Explanation:

For any chemical reaction to be spontaneous or to move in forward direction the ΔG ,the Gibbs free energy must be negative.

The cell potential of a battery is positive for a spontaneous reaction, so for a battery to give output its cell potential must be positive.

Thermodynamics and electro-chemistry are related in the following manner:

ΔG=-nFE

n=number of electrons involved

F=Faradays constant

E=cell pottential of battery

so from the above equation ΔG would only be negative when E cell that is the cell potential is positive.

For a battery which is being used its cell potential is positive and hence the ΔG would be negative. So the cell reaction occurring would be in forward direction as ΔG is negative.

when the cell potential Ecell is 0 then ΔG is also zero then the reaction occurring in battery would be at equilibrium.

When the cell potential Ecell is - then ΔG is positive and the reaction would be occurring backwards.

3 0
3 years ago
What atomic or hybrid orbitals make up the sigma bond between ge and h in germanium hydride, geh4 ?
ozzi
<span>Germane is the chemical compound with the formula GeH₄, and the germanium analogue of methane. It is the simplest germanium hydride and one of the most useful compounds of germanium.


</span>In chemistry, sigma bonds (σ bonds) are the strongest type of covalent chemical bond. They are formed by head-on overlapping between atomic orbitals. Sigma<span> bonding is most simply defined for diatomic molecules using the language and tools of symmetry groups.
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
8 0
3 years ago
Read 2 more answers
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