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shepuryov [24]
3 years ago
10

Write the rate law for the reaction of S2O8^2-

Chemistry
1 answer:
Tamiku [17]3 years ago
7 0

the rate law should be k[S2O8^-2]

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How many molecules are in 85g of silver nitrate?
maksim [4K]
<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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>Atomic Structure</u>

  • Reading a Periodic Table
  • Writing Compounds
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})
  2. Multiply/Divide:                                                                                                \displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0
3 years ago
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
HELP ASAP 20 POINTS AAAA
pochemuha

Answer:

Me and my friends were going to do a science experiment. Jonny’s job was to make the HYPOTHESIS. He said the “ If we mix baking soda and vinegar together, the TEMPERATURE will go down.”

So then Molly mixed the baking soda and vinegar together and checked the TEMPERATURE. We all OBSERVED as the thermometer’s TEMPERATURE went down. “ your THEORY/ HYPOTHESIS was correct!” Exclaimed Molly.

Then the whole science GROUP let out with a cheer! And wrote the information down on their EXPERIMENTAL info chart. They took a microscope and looked at the mixture because they wanted to the the little PARTICLES in the mixture. Lily CONTROLED the microscope she zoomed in and out to see the particles.

Explanation:

i hope this helps:)

3 0
2 years ago
You react 2.33 g of iron (III) chloride with 50.0 mL of 0.500 M solution of sodium phosphate to
nikklg [1K]

Answer:

2.33g of iron (iii) chloride

50.0 mL of 5.00 M of sodium phosphate

FeCl3 + Na3PO4 > Fe(PO4) + 3NaCl

mol = conc × vol = 0.5 × 50/1000 = 0.025 mol Na3PO4

from the equation:

1 mol of Na3PO4 reacts with 1 mol FeCl3 = 3 mol of NaCl

0.025 mol = x

x = 0.0025 × 3 = 0.075 mol NaCl

mass = 0.075 g × 59 g/mol = 4.425 g NaCl

i guessed all of this so i dont know i it is correct

3 0
2 years ago
Which of these is a chemical property of a substance?
dlinn [17]

Answer:

Reactivity

Explanation:

8 0
2 years ago
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