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lawyer [7]
3 years ago
5

Why do solids expand on heating?

Chemistry
1 answer:
sdas [7]3 years ago
5 0

Explanation:

The molecules of solids are shrinked in there normal state . but as a heat energy is produced , the molecules starts curating fast and fast as temperature goes up . since they vibrate , they hit and collide each other breaking the bondings this increases the surface of area of the solid , and molecules consumes that space and they expand .

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Kp for the following reaction is 0.16 at 25 degree C. 2 NOBr(g) 2 NO(g) Br_2(g) The enthalpy change for the reaction at standard
finlep [7]

Answer:

Explanation:

Given that:

2 NOBr_{(g)} \iff 2 NO_{(g)} + Br_{2(g)}

From above:

K_p = 0.16 = \dfrac{(P_{NO})^2 (P_{Br})}{(P_{NOBr})^2}

To predict the effect of the addition of Br₂(g);

The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr

The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.

5 0
2 years ago
In a chemical equation, changing the subscript in a chemical formula for a substance changes what aspect?
Citrus2011 [14]
The subscript in a chemical formula is the number written next to the element at the bottom part. For example, the chemical formula of water is H₂O. The subscript of H is 2, while the subscript of O is 1. The subscript represents the number of a certain element in one particle of the compound. So, if you change the subscript, you also change the number of a certain element per compound. In other words, you change the ratio.
4 0
3 years ago
Read 2 more answers
Where do you find an elements energy levels
ahrayia [7]
You can find an element's amount of energy level by determining their place on the periodic table. An element's amount of energy levels are represented by which period/ row they are in. For example, Calcium has 4 energy levels. I know this because it is in the fourth period on the table.

Hope this helps!
8 0
3 years ago
A drop of water with a mass of 0.48 g is vaporized at 100 ∘C and condenses on the surface of a 55- g block of aluminum that is i
bagirrra123 [75]

The final temperature in Celsius of the metal block is 49°C.

<h3>How to find the number of moles ?</h3>

Moles water = \frac{\text{Given mass}}{\text{Molar Mass}}

                     = \frac{0.48\ g}{18\ \text{g/mol}}

                     = 0.0266 moles  

                   

Heat lost by water = 0.0266 mol x 44.0 kJ/mol

                                = 1.17 kJ

                                = 1170 J           [1 kJ = 1000 J]

Heat lost = Heat gained

Heat gained by aluminum = 1170 J  

1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242  

1170 + 1242 = 49.7 T  

T = 48.5°C (49°C at two significant figures)

Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.

Learn more about the Moles here: brainly.com/question/15356425

#SPJ1

7 0
2 years ago
12. What is the volume of 0.07 mol of neon gas at STP?
scoundrel [369]
<h3>Answer:</h3>

2 L Ne

<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>

  • Using Dimensional Analysis
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

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

STP - 22.4 L per mole

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})
  2. Multiply:                              \displaystyle 1.568 \ L \ Ne

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

1.568 L Ne ≈ 2 L Ne

7 0
3 years ago
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