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3 years ago

What were elements characterized by in the 1860's

Chemistry

1 answer:

ahrayia [7]3 years ago

4 0

Answer:

Döbereiner singled out triads of the elements lithium, sodium and potassium as well as chlorine, bromine and iodine. He noticed that if the three members of a triad were ordered according to their atomic weights, the properties of the middle element fell in between those of the first and third elements.

Explanation:

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A sample of natural rubber (200.0 g) is vulcanized, with the complete consumption of 4.8 g of sulfur. Natural rubber is a polyme

nasty-shy [4]

Answer: 1.3% many crosslinks as isoprene units,

Explanation:

Given:

mass pf natural rubber= 200.0g

mass of sulphur = 4.8g

molar mass of sulphur =32g/mol

molar mass of isoprene = C5H8=( 12x5) +(1x8)= 68g/mol

Solution: we first find no of moles present in each using

no of moles = $\frac{mass}{molarmass}$

Isoprene: 200.0g x [1mole / 68g] = 2.94moles.

Sulfur: 4.8g x [1mole / 32g] x [1 mole crosslinks / 4 moles S] = 0.0375 moles crosslinks.

to find % crosslinked units, we have

0.0375 / 2.94 = 1.3% as many crosslinks as isoprene units,

3 0

3 years ago

El Oro es metal o no metal

slega [8]

No metal, es un mineral.

4 0

3 years ago

Read 2 more answers

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-2}M^{-2}s^{-1}$

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

3 0

3 years ago

Why did Dalton think it was important to use his system of symbols for the chemical elements?

Dimas [21]

Explanation:

to remind people that atoms were

4 0

4 years ago

Identify the process in which the entropy increases. Identify the process in which the entropy increases. the phase transition f

diamong [38]

Answer:

The phase transition from a solid to a gas.

Explanation:

Entropy is related to the moles of gases available. The more gaseous moles, the higher the entropy.

Identify the process in which the entropy increases.

the phase transition from a solid to a gas. YES. There are more gaseous moles available so entropy increases.
a decrease in the number of moles of a gas during a chemical reaction. NO. This causes a decrease in entropy.
the phase transition from a liquid to a solid. NO. There is almost no change in the entropy in this process.
the phase transition from a gas to a liquid. NO. There is a decrease in entropy in this process.
the phase transition from a gas to a solid. NO. There is a decrease in entropy in this process.

7 0

4 years ago