Answer:
K, the rate constant = 9.73 × 10^(-1)/s
Explanation:
r = K × [A]^x × [B]^y
r = Rate = 1.07 × 10^(-1)/s
K = Rate constant
A and B = Concentration in mol/dm^-3
A = 0.44M
B = 0.11M
x = Order of reaction with respect to A = 0
y = Order of reaction with respect to B = 1
Solving, we get
r/([A]^x × [B]^y) = K
K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727
K = 0.9727
Answer:
True.
Explanation:
Here is an example: Hubble Space Telescope's launch in 1990 sped humanity to one of its greatest advances in that journey. Hubble is a telescope that orbits Earth. Its position above the atmosphere, which distorts and blocks the light that reaches our planet, gives it a view of the universe that typically far surpasses that of ground-based telescopes.
Hubble is one of NASA's most successful and long-lasting science missions. It has beamed hundreds of thousands of images back to Earth, shedding light on many of the great mysteries of astronomy. Its gaze has helped determine the age of the universe, the identity of quasars, and the existence of dark energy.
When a specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state, this emitted energy can
<span>be used to determine the "identity of the element".</span>
The elements in which electrons are progressively filled in 4f-orbital are called lanthanoids.
There are 15 metallic chemical elements called lanthanoids.
Those elements are with atomic numbers from 57 to 71.
Lanthanoids belong to f-block elements because of the filling of the 4f-orbitals.
One of the lanthanoids, lutetium (chemical symbol Lu) has 14 electrons if 4f orbitals.
Electron configuration of lutetium: [Xe] 4f14 5d1 6s2
f-orbitals are similar to the d orbital, but cut in half (eight lobes instead of four).
Lanthanoids are used not as materials in catalysts, alloys, lasers and cathode-ray tubes.
More info about lanthanoids: brainly.com/question/24413965
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Answer:
Explanation:
The missing image is attached below.
The objective of this question is to draw the major product formed from the diagram attached below.
From the diagram attached, we will see the reaction of a tertiary alkyl halide together with a weak nucleophile (ch3ch2oh) undergoing a nucleophilic substitution (SN₁) mechanism to yield a racemic mixture(i.e., compound that is not optically active but contains an equal amount of dextrorotatory and levorotatory stereoisomers) as a product.
