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AfilCa [17]
3 years ago
15

A) Devices that transfer kinetic energy have what for their source of power?

Chemistry
1 answer:
GarryVolchara [31]3 years ago
4 0

Answer:

A) Devices that transfer kinetic energy have a source of power that is in motion

Kinetic energy is the energy in motion, as such, a device that transfers kinetic energy transfers the energy the power source has into other energy forms

B) Kerosene does not easily cold start like diesel which can burn after compression

C) The first law of thermodynamics states that energy is conserved and it can neither be created nor destroyed, but can be changed from one form to another.

Therefore, when energy is not available in a given location or body, it cannot be obtained from that body or location

Explanation:

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If the mass of reactants equals 100g, then the mass of the products would be equal to_______?
ioda
The answer would be A, 100g since it’s equal
7 0
2 years ago
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How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -&gt;<br> 4A1 + 302
Evgesh-ka [11]
<h3>Answer:</h3>

\displaystyle 40 \ mol \ Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 40 \ mol \ Al

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0
3 years ago
Chemistry gcse
mart [117]
Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.

Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)
8 0
3 years ago
Select the correct scientist for each of the following advances in creating the modern periodic table.
sergey [27]

Answer:

Lavoisier; Newlands; Moseley

Explanation:

In 1789, Antoine Lavoisier grouped the elements into gases, nonmetals, metals, and earths.

In 1865, John Newlands developed the Law of Octaves. He stated that "any given element will exhibit analogous behaviour to the eighth element following it in the table."  

In 1914, Henry Moseley found a correlation between the X-ray wavelength of an element and its atomic number. He was then able to restructure the periodic table according to atomic numbers.

3 0
3 years ago
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Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

5 0
3 years ago
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