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Katyanochek1 [597]
3 years ago
8

How many milliliters of water will be created from a combustion reaction with 9.32×10 22nd power of ethanol molecules. Assume de

nsity of water is 1.00 g/ml and molar mass of water is 18.02 g/mo
Chemistry
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

8.38 mL

Explanation:

1- The combustion reaction of ethanol (C2H5OH) in the presence of oxygen (O2) has as reaction products carbon dioxide (CO2) and water (H2O), the equation equaled is:

<em>C2H5OH (l) + 3 O2 (g) CO2 2 CO2 (g) + 3 H2O (l) </em>

2- By establishing the stoichiometric relationship between ethanol and water, you can calculate the number of molecules that will be created from the initial amount of alcohol molecules:

6,022x10 23 molecules of C2H5OH (1 mol) ___ 3 x 6,022x10 23 molecules of H2O (3 moles)

9.32x10 22 C2H5OH molecules _____ X = 2.80x10 23 H2O molecules

<em>Calculation: </em>

9.32x10 22 x (3 x 6.022x10 23) / 6.022x10 23 = 2.80x10 23 H2O molecules

3- Once the number of water molecules formed is obtained, with the molar mass the mass can be determined:

6.022x10 23 H2O molecules _____ 18.02 g

2.80x10 23 molecules of H2O _____ X = 8.38 g of H2O

<em>Calculation: </em>

2.80x10 23 x 18.02 g / 6.022x1023 = 8.38 g of H2O

4- Finally, having the density of water, you can calculate the volume that formed:

d = m / V  --> V = m / d

V = 8.38 g / 1.00 mL = 8.38 mL

The answer is that 8.38 mL of water is formed

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worty [1.4K]

Answer:

The correct statement is:

E - The entropy of the products is greater than the entropy of the reactants.

Explanation:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

As glucose is a large molecule and then it is transformed into many molecules of water and carbon dioxide, the entropy of the system increases. If the number of molecules increases, the disorder increases.

Initial state: 7 molecules (1 glucose + 6 oxygen)

Final state: 12 molecules (6 carbon dioxide + 6 water)

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3 years ago
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A particular atom contains 29 electrons, 34 neutrons, and 29 protons. what is the identity of this element and what is its atomi
IgorLugansk [536]

Answer:

             Identity =  Copper

             Atomic Number =  29

Explanation:

Given Data:

Number of Electrons  =  29

Number of Neutrons =  34

Number of Protons  =  29

Identity and atomic number of Element:

                              This element can be identified by using two sources, i) Atomic Number ii) Electronic Configuration.

Atomic Number is defined as the number of protons present in the nucleus of an element. As the given number of protons are 29, therefore, the atomic number of this element is also 29.

Secondly, if the element is in neutral state then its number of protons must be equal to the number of electrons present in it. So, in neutral state this element will contain 29 electrons. Therefore, the electronic configuration is as follow,

                                     1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁹

Now using electronic configuration one can easily identify the position of this element in periodic table. As the valence shell is 4, so it belongs to period 4 and there are 9 electrons in d-subshell so it belongs to transition elemnts and transition element with 9 electrons in d shell is Copper. Also, the stable electronic configuration of copper is written as,

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3 years ago
Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or nega
algol [13]

Answer: a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s):  negative

b)  CaCO_3(s)\rightarrow CaO(s)+CO_2(g) : positive

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g): positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s) : negative

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g):  positive.

f) I_2(s)\rightarrow I_2(g) : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

As ions are moving to solid form , randomness decreases and thus sign of \Delta S is negative.

b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g)

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of \Delta S is positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)

As gas is changing to solid, randomness decreases and thus sign of \Delta S is negative.

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of \Delta S is positive.

f) I_2(s)\rightarrow I_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

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Answer:

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Explanation:

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