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VikaD [51]
3 years ago
5

Please help!! I don’t understand

Chemistry
1 answer:
laiz [17]3 years ago
4 0
<h3>Answer:</h3>

150 g Si

<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
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[Given] 3.2 × 10²⁴ atoms Si

[Solve] grams Si

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

Avogadro's Number

[PT] Molar Mass of Si - 28.09 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.2 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})(\frac{28.09 \ g \ Si}{1 \ mol \ Si})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 149.266 \ g \ Si

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. Instructed to round to 2 sig figs.</em>

149.266 g Si ≈ 150 g Si

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This reaction supports the Law of Conservation of Mass (hint: is the equation balanced?):
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False

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<em>Left:</em>

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O = 1 atoms

H = 1 atoms

<em>Right: </em>

K = 2 atoms

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H = 2 atoms

O = 1 atoms

As you can see both sides do not have the same amount of atoms, therefore it is not balanced.

<h3>Hope this Helps!!</h3>

:)

7 0
3 years ago
Can someone help me out please ? I am very stuck here
Alex17521 [72]
Hey there!

<span>Atomic Masses :
</span>
H = <span>1.00794  a.m.u
N = </span><span>14.0067  a.m.u
O = </span><span>15.9994  a.m.u

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Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

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