El átomo de bromo tiene Z=35 la configuración electrónica por niveles y subniveles de energía

Thepotemich [5.8K]

Answer:

1-Pentene

Explanation:

If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.

For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.

Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.

For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.

3 0

2 years ago

The balanced equation for a hypothetical reaction is a + 5b + 6c → 3d + 3e. what is the rate law for this reaction? rate = k [a]

nekit [7.7K]

<u>Answer: </u>The correct rate of the reaction is $Rate=k[a][b]^5[c]^6$

<u>Explanation:</u>

Rate law of the reaction is the expression which expresses the rate of the reaction in the terms of the molar concentrations of the reactants with each term raised to the power of their respective stoichiometric coefficients in a balanced chemical equation.

For the given reaction:

$a+5b+6c\rightarrow 3d+3e$

The expression for the rate law will be: $Rate=k[a][b]^5[c]^6$

7 0

3 years ago

1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm

sashaice [31]

1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm

Ideal gas equation: pV = nRT => n = pV / RT

R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm

=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol

2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type

X (+) + O2 (g) ---> X2O or

2 X(2+) + O2(g) ----> X2O2 = 2XO or

4X(3+) + 3O2(g) ---> 2X2O3

In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)

So, lets probe those 3 cases.

3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X

Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol

That does not correspond to any of the metal with valence 1+

So, now probe the case 2.

4) Case 2:

2moles X metal / 1 mol O2(g) = x / 0.01226 mol

=> x = 2 * 0.01226 = 0.02452 mol

And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol

That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.

4) Case 3

4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635

atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol

That does not correspond to any metal.

Conclusion: the identity of the metallic element could be titanium.

5 0

3 years ago

T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,

azamat

Answer:

TRUE
FALSE
TRUE
FALSE
FALSE
TRUE

Explanation:

At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
Faster solidification results in smaller interlamellar spacing : TRUE

4 0

2 years ago

A) 30.22 g NaCl x 1 mol NaC l = 58.4430 Molar mass (g) NaCl=0.5171 mol NaCl

jekas [21]

The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.

So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl.

6 0

2 years ago