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Diano4ka-milaya [45]
3 years ago
12

I what process does a solid change directly into a vapor​

Chemistry
1 answer:
Marina86 [1]3 years ago
7 0

Answer:

solid change directly into a vapor in the process known simply as freezing

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you just got home from a run on a hot Atlanta afternoon. you grab a 1.00-liter bottle of water and drink three-quarters of it in
nikitadnepr [17]

Answer:

41.67 mol

Explanation:

1 Litre of water = 1000g

Mole = mass / molar mass

Mass of 1 L of water = 1000 g

Molar mass of water (H2O) :

(H = 1, O = 16)

H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol

Amount of water consumed = (3/4) of 1 litre

= (3/4) * 1000g

= 750g

Therefore mass of water consumed = 750g

Mole = 750g / 18g/mol

Mole of water consumed = 41.6666

= 41.67 mol

3 0
3 years ago
How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?
zlopas [31]
<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<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>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

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

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. Multiply:                            \displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0
3 years ago
What was the key design change for hfc-134a a/c systems versus CFC 12 a/c systems
BabaBlast [244]
Answer is: key design change for HFC-134a A/C systems versus CFC-12 A/C systems was quick couple service fitting and that design reduce venting and mixing of refrigerants during service.

<span> Level of contamination is also reduced and the emission of refrigerants and greenhouse gases (sulfur dioxide, carbon dioxide) is also reduced.</span>

6 0
3 years ago
A piston is pressurized to 15.5 psi at 405 K. If the piston compresses the air, from 8.98 L to 7.55 L, and the pressure drops to
masya89 [10]

Answer:

239.45 K

Explanation:

Ideal gas law formula is P1V1T2=P2V2T1

Rearrange that to get...

T2=T1P2V2/P1V1

Fill in the values and solve.

7 0
3 years ago
Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio
soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂  → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • CuCl₂: 1 mole
  • CuCO₃: 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • Na₂CO₃: 129 g/mole
  • CuCl₂: 134.45 g/mole
  • CuCO₃: 123.55 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Na₂CO₃: 1 mole ×129 g/mole= 129 grams
  • CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
  • CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
  • NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
1 year ago
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