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Nonamiya [84]
4 years ago
5

Conditions for an experimental chemistry reaction require a temperature of 300 K. The temperature in the lab is 55°F. Which of t

he following must you do to meet the requirements? (°F= 9/5 (K - 273) + 32
Chemistry
2 answers:
Mekhanik [1.2K]4 years ago
8 0
You just need to put in K = 300 and solve for F.

F = 9/5 (300 - 273)
F = 9/5 (27)
F = 48.6 °

if the room is 55°F the temperature will need to be turned down to get to 48.6°F
Marina86 [1]4 years ago
7 0

Answer:

 

decrease the room temperature by 33°F

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3 years ago
How many molecules are in 0.55 moles of Cu(NO3)2?
xeze [42]
<h3>Answer:</h3>

3.3 × 10²³ molecules Cu(NO₃)₂

<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>

  • 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>

0.55 mol Cu(NO₃)₂

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.55 \ mol \ Cu(NO_3)_2(\frac{6.022 \cdot 10^{23} \ molecules \ Cu(NO_3)_2}{1 \ mol \ Cu(NO_3)_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 3.3121 \cdot 10^{23} \ molecules \ Cu(NO_3)_2

<u>Step 4: Check</u>

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

3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂

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C5H12() + 502(9)=
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The correct answer is B
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What's the oxidation # of P in Mg3p2
ale4655 [162]
The oxidation number of P in Mg3P2 is 3. When writing ionic compounds, you swap the oxidation numbers and add them as the subscript. 
5 0
3 years ago
How many grams of KNO3, are needed to create 675.0 mL with a concentration of 1.71 M
valkas [14]

Answer:

Бардык белгилүү авиация мыйзамдарына ылайык, аары учуу мүмкүнчүлүгү жок. Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.

Explanation:

Канаттары өтө эле кичинекей, семиз денесин чече албайт. Албетте, аары баары бир учат. Себеби аарылар адам мүмкүн эмес деп эсептеген нерсеге маани бербейт.

4 0
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