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Marizza181 [45]
3 years ago
7

Sediment layers stop lateral spreading when:

Chemistry
2 answers:
Andrew [12]3 years ago
6 0

Answer:

they encounter a barrier and they run out of additional sedimentary material

Explanation:

GREYUIT [131]3 years ago
3 0
Sediment layers stop lateral spreading when they encounter a barrier and they run out of additional sedimentary material. Lateral spreading is the lateral movement of gently to steeply sloping, saturated soil deposits caused by earthquake-induced liquefaction. Hope this answers the question.
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Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.(b) Cite two important additional refi
Alenkinab [10]

Answer: Bohr postulated that electronic energy levels are quantized. Secondly, a photon of light of a particular frequency is emitted when electrons move from a higher to a lower energy levels.

Explanation:

The Bohr model of the atom is the immediate predecessor of the wave mechanical model of the atom. The wave mechanical model refined the Bohr's model by treating the electron as a wave having a wave function psi. The wave function describes the identity of the electron. From Heisenberg uncertainty principle, the position of a particle cannot be accurately and precisely measured. Hence the wave mechanical model added that electrons are not localized in orbits according to Bohr's model but the integral of psi squared dx gives the probability of finding the electron within a given space.

8 0
3 years ago
Read 2 more answers
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
Decreasing the temperature of the reaction 3H2 + N2 2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL THE REACTIO
Alex

Answer:

the reaction will shift towards the “heat”—shifts to the left

Explanation:

To summarize:

o If temperature increases (adding heat), the reaction will shift away from the “heat” term and go in the

endothermic direction.

o If temperature decreases (removing heat), the reaction will shift towards the “heat” term and go in the

exothermic direction.

o NOTE: The endothermic direction is always away from the “heat” term and the exothermic direction is

towards the “heat” term.

Therefore the reaction will shift towards the “heat”—shifts to the left

3 0
3 years ago
11) A sample of rhenium of suspected extrasolar origin (translation: not from this solar
jarptica [38.1K]

<u>Answer:</u>

<em>Atomic number 75 is dedicated to an element named rhenium and has been given Re as its chemical name.</em>

<u>Explanation:</u>

With a really low concentration it is one of the rarest metals that is found in Earth's crust.

Like all other elements rhenium also has certain isotopes among with 185 and 187 are the most stable ones. Hence these two are the ones that are naturally available abundance is 34% and 63% respectively.

4 0
3 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
3 years ago
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