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aniked [119]
3 years ago
8

If the products in a chemical equation include h2o, what specific type of reaction is occuring? Is a polymer being made or broke

n down in this equation?
Chemistry
1 answer:
anzhelika [568]3 years ago
7 0

There are two kinds of polymers : addition polymers and condensation polymers

In case of addition polymers, there is simple addition of unsaturated monomeric units. example: polythene

While in case of condensation polymers, the monomers combine with the elimination of a small molecule like H2O, HCl, NH3 etc.

So here as there is formation of H2O as product there is formation of a polymer and that too a condensation polymer

Example: condensation polymer of diamine and dicarboxylic acid to form a polyamide (Nylon-6,6)

Answer: condensation

formation of polymer

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Answer : The correct answer is, Halide hydrocarbon.

Explanation :

Freon : Freon are the simple aliphatic organic compounds which may contains fluorine, carbon, hydrogen, chlorine or bromine. So, they are a type of chlorofluoro carbons (CFCs).

They are considered as refrigerants which are used in air-conditioning and refrigeration system.

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Which is the command center of the body​
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nervous system

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The maximum depth of Lake Tahoe is 99.6 rod. How deep is Lake Tahoe in fathoms if one fathom-0.364 rod? ball part 250-300 fathom
Romashka [77]

Answer:

273.9 fathom

Explanation:

Fathom (ftm) and rod are the units of length used in the imperial and the U.S. customary systems.

Fathom is an unit used for the measurement of depth of water. It is equal to 1.8288 m.

Rod, also known as perch or pole, is equal to 5.0292 m.

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and 1 rod = 2.75 fathom

∴ 99.6 rod = 99.6 × 2.75 fathom = 273.9 fathom

<u>Therefore, maximum depth of Lake Tahoe = 99.6 rod = 273.9 fathom</u>

6 0
3 years ago
The magnesium atom in chlorophyll A. is used to pass excited electrons on to pheophytin B. is coupled to the production of ATP C
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Consider the following intermediate chemical equations.
Zinaida [17]

Answer:

The first choice, -205.7\; \rm kJ.

Explanation:

Let the three reactions, where the enthalpy change were known, be called (1), (2), and (3).

The goal is to find the enthalpy change of the fourth equation. Assume that this equation can be written as x \times (1) + y \times (2) + z \times (3) for some x, y, and z (might not be whole numbers or take positive values.) Then, by Hess's Law, the enthalpy change of that reaction would be x \cdot \Delta H_1 + y \cdot \Delta H_2 + z \cdot \Delta H_3.

To find these x, y, and z, consider: what combination of reaction (1), (2), and (3) would give the fourth reaction?

Imagine that the coefficients are positive for all the reactants, and negative for all the products.

For example: in (1), \rm H_2\; (g) has a coefficient of 2. However, since it is on the the product side of (1), its value should be -2. Also, in (3)

Since there is no \rm H_2\; (g) in the desired equation, the value of x, y, and z should ensure that -2x + z = 0.

Another example: \rm CH_4\; (g) is on the reactant side of the first reaction. Its coefficient in the equation is 1, so that corresponds to +1. Since \rm CH_4 is neither in (2) nor in (3), the value of

In the desired equation, \rm CH_4\; (g) is on the reactant side with a coefficient of 1. As a result, the value of x, y, and z should ensure that x = 1.

One such equation can be found for each species in the reactions.

\displaystyle \begin{cases}x = 1 & \left(\text{For $\mathrm{CH_4\; (g)}$}\right) \\ -x - y= 0 & \left(\text{For $\mathrm{C\; (s)}$}\right) \\ -2\, x + z = 0 & \left(\text{For $\mathrm{H_2\; (g)}$}\right) \\ y = -1 & \left(\text{For $\mathrm{CCl_4\; (g)}$}\right) \\ -2\, y  + z= 4 & \left(\text{For $\mathrm{Cl_2\; (g)}$}\right) \\  -2\, z = -4 & \left(\text{For $\mathrm{HCl\; (g)}$}\right)\end{cases}.

Solve this system of equations for x, y, and z (this approach works only if at least one solution exists.) In this case,

\displaystyle \begin{cases}x = 1 \\ y = -1 \\ z = 2\end{cases}.

Calculate the enthalpy change of the desired reaction:

\begin{aligned}\Delta H &= x\times \Delta H_1 + y \times \Delta H_2 + z \times \Delta H_3 \\ &= 1 \times 74.6 + (-1) \times 95.7 + 2 \times (-92.3) \\ &= -205.7\; \rm kJ\end{aligned}.

4 0
3 years ago
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