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Neporo4naja [7]
2 years ago
9

Can someone please do the last column for me

Chemistry
1 answer:
JulsSmile [24]2 years ago
8 0

Answer:

First one is : London dispersion

Second one is: London dispersion

Third One is : dipole-dipole

Fourth one is : induced dipole

Explanation:

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A single atom of an element has 11 protons, 11 electrons, and 12 neutrons. Which element is it?
nataly862011 [7]

Answer:

Na

Explanation:

8 0
3 years ago
What is the mass of 1.58 moles of CH4
HACTEHA [7]
<h3>Answer:</h3>

25.4 g CH₄

<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

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

1.58 mol CH₄

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

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})
  2. Multiply/Divide:                 \displaystyle 25.359 \ g \ CH_4

<u>Step 4: Check</u>

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

25.359 g CH₄ ≈ 25.4 g CH₄

4 0
3 years ago
1. the mass of one electron is
kupik [55]

Answer:

Mass of one electron is 9.1 × 10⁻³¹ kg

Mass of one proton is 1.673 × 10⁻²⁷ Kg

Mass of one neutron is 1.675 × 10⁻²⁷ Kg

 

<u>-TheUnknownScientist</u><u> 72</u>

4 0
2 years ago
Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.
Ronch [10]

Answer : The limiting reagent is O_2

Solution : Given,

Moles of methane = 2.8 moles

Moles of O_2 = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 5 moles of O_2 react with \frac{5}{2}=2.5 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2

8 0
3 years ago
Under ordinary conditions of temperature and pressure, the particles in a gas are
vova2212 [387]
<span>In normal conditions gas particles remain very distant from each other. They rarely collide and are stable. When temperature increases the gas particles begin to move faster and collide more, reducing the distance. When pressure increases the gas particles also pick up kinetic speed and are also closer to each other.</span>
4 0
3 years ago
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