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Maru [420]
3 years ago
15

40.0 L of oxygen were collected at 10°C and 758 mm of Hg. Calculate its volume at STP.

Chemistry
1 answer:
marta [7]3 years ago
3 0

Answer:

189.5

Explanation:

its very very very very easy

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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the
kramer

Answer:

The equilibrium constant of the given reaction is 0.01351.

Explanation:

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

The equilibrium constant of the reaction = K_3=1.85\times 10^{-10}

K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}...[1]

AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)

The equilibrium constant of the reaction = K_4=1.17\times 10^{-4}

K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}..[2]

[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}

PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)

The expression of equilibrium constant of the creation is ;

K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}

Dividing [1] by [2]

\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}

\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}

Substituting the value of [Cl^-] from [2] :

\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}

\frac{K_3}{K_4}=K_4\times K

K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}

K=0.01351

The equilibrium constant of the given reaction is 0.01351.

3 0
3 years ago
What atomic structure has 56 protons and 84 neutrons?
Nikitich [7]
Name:<span> Polonium </span>
Symbol:<span> Po </span>
Atomic Number:<span> 84 </span>
Atomic Mass:<span> (209.0) amu </span>
Melting Point:<span> 254.0 °C (527.15 K, 489.2 °F) </span>
Boiling Point:<span> 962.0 °C (1235.15 K, 1763.6 °F) </span>
Number of Protons/Electrons:<span> 84 </span>
Number of Neutrons:<span> 125 </span>
Classification: Metalloid 
Crystal Structure:<span> Monoclinic </span>
Density @ 293 K:<span> 9.4 g/cm</span>3 
Color:<span> Unknown</span>


Atomic Structure<span> <span>
Number of Energy Levels: 6<span>
First Energy Level: 2 
Second Energy Level: 8 
Third Energy Level: 18 
Fourth Energy Level: 32 
Fifth Energy Level: 18 
Sixth Energy Level: 6
does it help?</span></span></span>
3 0
4 years ago
What is one main role of state governments?
mr Goodwill [35]
C.commanding the army the the navy would be right I think
7 0
2 years ago
Read 2 more answers
How likely do you think it is that life exists on other planets
mel-nik [20]

Answer:

not very likely

Explanation:

3 0
3 years ago
Given: 36.7 grams of CaF2 is added to 300 mL water. Find molarity?
BigorU [14]
<h3>Answer:</h3>

2 M

<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>Chemistry</u>

<u>Unit 0</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<u>Aqueous Solutions</u>

  • Molarity = moles of solute / liters of solution
<h3>Explanation:</h3>

<u>Step 1: Define</u>

36.7 g CaF₂

300 mL H₂O

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

Molar Mass of Ca - 40.08 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol

1000 mL = 1 L

<u>Step 3: Convert</u>

<em>Solute</em>

  1. Set up:                               \displaystyle 36.7 \ g \ CaF_2(\frac{1 \ mol \ CaF_2}{78.08 \ g \ CaF_2})
  2. Multiply:                             \displaystyle 0.470031 \ mol \ CaF_2

<em>Solution</em>

  1. Set up:                              \displaystyle 300 \ mL \ H_2O(\frac{1 \ L \ H_2O}{1000 \ mL \ H_2O})
  2. Multiply:                            \displaystyle 0.3 \ L \ H_2O

<u>Step 4: Find Molarity</u>

  1. Substitute [M]:                    \displaystyle x \ M = \frac{0.470031 \ mol \ CaF_2}{.3 \ L \ H_2O}
  2. Divide:                                \displaystyle x = 1.56677 \ M

<u>Step 5: Check</u>

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

1.56677 M ≈ 2 M

8 0
3 years ago
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