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RideAnS [48]
3 years ago
10

Circle only those of the following reactions that are redox reactions:

Chemistry
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer: options B,D and F

Explanation:

Since redox reactions are those which involves both oxidation and reduction

In B , Cu is oxidized and S gets reduced

D, Na gets oxidized and hydrogen gets reduced

F, carbon gets oxidized and Oxygen gets reduced

In g, there is no change in oxidation no of s in both product and Reactants is same +4

Similarly in the case of Ag and Mg.

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... in neutralization reactions, for example:

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Explain the role of each factor in chemical weathering (water, acid, and air).
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Answer: hydrolysis oxidation acidification

Explanation:

they break down, dissolve, or create new elements

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Calcium carbon oxygen, calcium chlorine , carbon hydrogen oxygen, hydrogen chlorine, calcium oxygen hydrogen. pretty sure they’re right
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3 years ago
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What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?
Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})

Hence, the probability of finding the particle in the  one-dimensional box is:

P = \int_{x_{1}}^{x_{2}} \psi^{2} dx

P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx

P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]

P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})    

Solving for n=4:

P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})    

P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})    

P = \frac{1}{8}

I hope it helps you!

7 0
3 years ago
One of relatively few reactions that takes place directly between two solids at room temperature is <img src="https://tex.z-dn.n
kolbaska11 [484]

Answer:

a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) 3.14g must be added

Explanation:

a) For the reaction:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)

To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:

6.5 g × (1mol / 315.48g) =<em> 0.0206moles of  Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:

0.0206moles of  Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:

0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>

8 0
3 years ago
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