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Sergeu [11.5K]
3 years ago
8

Which law states that at constant temperature the pressure of a given amount of gas is inversely proportional to the volume of t

he gas?
Chemistry
1 answer:
Phantasy [73]3 years ago
8 0
That law is known as Boyle's Law, "The volume of a given mass of a gas is inversely related to pressure when the temperature is constant"
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1. If you have 25% radioactive parent atoms in a mineral crystal, and 75% daughter atoms, how many half-lives have passed?
Andreyy89
Too freaking many... or maybe not many at all

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3 years ago
Which colorless and odorless gas, produced by radioactive decay of Uranium-238, is considered to be a cancer-causing agent?
alexdok [17]
<span>The correct answer is b. Radon. Oxygen, Hyrdrogen, and Boron, are not radioactivel ike Radon and it regenerates quickly meaning that even though it has a short half-life period, it stays for a long time once released. It also has no taste so it's difficult to notice without proper gear because you can't feel it.</span>
7 0
4 years ago
A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6
Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

H = \frac{E}{M*(T2-T1)}

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}

Finally, the specific heat of this metal is

H = 993 \frac{J}{kgC}

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally,  we can conclude that this metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

7 0
3 years ago
LOS AMOOOO 1 AL 40 COMENTÉN​
OleMash [197]

Answer:

0 ni siquiera te conozco

Explanation:

6 0
3 years ago
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
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