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Advocard [28]
3 years ago
12

Lord kelvin described the concept of absolute zero temperature and the laws relating the change in thermal energy during chemica

l reactions. whaty type of chemist would be be considered todayh
Chemistry
1 answer:
kondaur [170]3 years ago
6 0
Lord Kelvin, were he alive today, would be considered a Thermochemist. Thermochemistry is interested in the role of heat in chemical reactions. This includes the role of heat both as a biproduct of chemical reactions and a facilitator.

Kelvin's description of absolute zero is an important concept in thermochemistry. At absolute zero, there is no movement of molecules, and no energy available facilitate chemical reactions. 
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How many grams are in 1 mole of Ar?
Ilia_Sergeevich [38]

Answer: 39.948 grams

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Ar, or 39.948 grams

3 0
3 years ago
If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?
bija089 [108]
<h3>Molar mass:-</h3>

\\ \sf\longmapsto HC_2H_3O_2

\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)

\\ \sf\longmapsto 1u+24u+3u+48u

\\ \sf\longmapsto 28u+48u

\\ \sf\longmapsto 76u

\\ \sf\longmapsto 76g/mol

  • No of moles=0.2mol
  • Given mass=?

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}

\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}

\\ \sf\longmapsto Given\:Mass=0.2\times 76

\\ \sf\longmapsto Given\:Mass=1.52g

7 0
3 years ago
1. How many protons does 14/6 C contain? What element is this?
Harman [31]

Answer:

It has 6 protons and its Carbon 14

Explanation:

3 0
3 years ago
Two isotopes of hypothetical element X exist with abundances of 30.00% 100X and 70.00% 101X. What is the approximate atomic mass
Rasek [7]

Answer:

C. 100.7 amu

Explanation:

Isotopes of an element are atoms of an element with the same atomic number but different atomic masses. Each atomic mass of an isotope is known as an isotopic mass. An element that exhibits isotope, that is, that have two or more isotopes has a relative atomic mass that is not a whole number.

Relative atomic mass of X is the sum of the products of the relative abundances of each isotope and its isotopic mass.

For Isotope ¹⁰⁰X: 30% × 100 = 30 amu

For Isotope ¹⁰¹X: 70% × 101 = 70.7 amu

Relative atomic mass of X = (30 + 70.7) amu = 100.7 amu

Therefore, the approximate atomic mass of X is 100.7 amu

4 0
3 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

6 0
3 years ago
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