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Nonamiya [84]
2 years ago
8

Of the bonds listed below the weakest bond is _______ and the strongest is ________. A strong bond is a (stable / unstable) bond

2(H—C ≡ C—H) 5(O = O) → 4 (O—C—O) 2 (H—O—H) 839 kJ/mol C ≡ C.
Chemistry
1 answer:
zhannawk [14.2K]2 years ago
6 0
Weakest C-H. Strongest: C = C.
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Why is cold air cold?
Sholpan [36]

Explanation:

Energy makes molecules in the air move faster and expand, decreasing density. The opposite is true for cold air. Molecules are closer together, and absorb less energy.

4 0
3 years ago
What is the engine piston displacement in liters of an engine whose displacement is listed as 490 in^3?
marishachu [46]

Answer:

490 in^3 = 8.03 L

Explanation:

Given:

The engine displacement = 490 in^3

= 490 in³

To determine the engine piston displacement in liters L;

(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)

First, we will convert in³ to cm³

Since 1 in = 2.54 cm

∴ 1 in³ = 16.387 cm³

If 1 in³ = 16.387 cm³

Then 490 in³ =  (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³

∴ 490 in³ = 8029.63 cm³

Now will convert cm³ to dm³

(NOTE: 1 L = 1 dm³)

1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm

∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³

If 1 cm³ = 1 × 10⁻³ dm³

Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³

≅ 8.03 dm³

∴ 8029.63 cm³ = 8.03 dm³

Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³

Since 1L = 1 dm³

∴ 8.03 dm³ = 8.03 L

Hence, 490 in³ = 8.03 L

3 0
2 years ago
sp2 hybrid orbitals have 1. trigonal bipyramidal symmetry. 2. linear symmetry. 3. tetrahedral symmetry. 4. trigonal planar symme
Cloud [144]

Answer:

4. trigonal planar symmetry.

Explanation:

The sp2 hybridization is formed with one s and two p atomic orbitals and form  trigonal planar symmetry. In sp2 hybridization, there are same valence shell in both the orbitals and it gives three equivalent sp2 hybridized orbitals that are separated by 120 degrees giving trigonal planar symmetry.

Hence, the correct answer is "4. trigonal planar symmetry.".

7 0
3 years ago
8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8
garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7)  kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

3 0
3 years ago
HELPPP LOOK AT PICTURE
lina2011 [118]

Answer:

it is a square

Explanation:

I hope it will helps you

sorry it's not a true answer

because I want points

5 0
3 years ago
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