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Lina20 [59]
3 years ago
10

Can anybody can solve this PLEASE

Chemistry
1 answer:
strojnjashka [21]3 years ago
8 0

Answer:

Explanation:

b is the most stable( noble gas ) since it has an octet valance shell and can't loses or gains any more of electrons

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different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity fr
SCORPION-xisa [38]

Answer:density

Explanation:

3 0
4 years ago
At what temperature (in K) will 3.5 moles of gas occupy 2.7 L at 1.5 atm? HELP PLEASE!
Vesnalui [34]

3.5 moles of a gas will occupy 2.7 L at 1.5 atm at a temperature of 14.1K

IDEAL GAS LAW:

  • The temperature of a gas can be calculated using the ideal gas law equation:

PV = nRT

Where;

  1. P = pressure (atm)
  2. V = volume (L)
  3. n = number of moles (mol)
  4. R = gas law constant (0.0821 Latm/molK)
  5. T = temperature (K)

  • According to this question, P = 1.5atm, V = 2.7L, n = 3.5moles, T = ?

  • 1.5 × 2.7 = 3.5 × 0.0821 × T

  • 4.05 = 0.28735T

  • T = 4.05 ÷ 0.28735

  • T = 14.1K

  • Therefore, 3.5 moles of a gas will occupy 2.7 L at 1.5 atm at a temperature of 14.1K

Learn more at: brainly.com/question/13821925?referrer=searchResults

6 0
3 years ago
Please help help help help
Anni [7]

Answer:

50000 dollars or 5 e5

Explanation:

Mr. Garibay has 5.0 * 10^4 dollars in his bank.

Scientific notation represents the data in the format of expanded number or with e character.

For instance, 5.0 * 10^4 dollars = 50000 dollars or 5 e5

3 0
3 years ago
HNO3 and H2CO3 are examples of ?
Basile [38]
A) acids because they start with h
7 0
3 years ago
When 2 grams of powdered lead (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Lead (IV) ox
kirza4 [7]

Answer:

The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

Explanation:

Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen  using Lead (IV) oxide as a catalyst.

  1. The catalyst surface area is directly proportional to the reaction rate
  • So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.

     2. Also, Removing lead (IV) oxide from the reaction mixture  the reaction rate decreased because as the catalyst is removed.

     3.  Using 50 cm³ of hydrogen peroxide  doesn't affect the rate because the concentration of the reactant doesn't change.

     4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased

So, The right answer is:

Replacing the powdered lead oxide with its large crystals

Removing lead (IV) oxide from the reaction mixture

Using 1.0 gram of lead (IV) oxide

7 0
4 years ago
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