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3 years ago

Who was the first person to propose that atoms existed?

Chemistry

2 answers:

Gennadij [26K]3 years ago

8 0

Democritus was the first person to theorize the existence of atoms.

levacccp [35]3 years ago

7 0

Remember we are just talking about proposals, not actual evidence. So it would be Democritus.

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

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

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

Expression for rate law for fourth observation:

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

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

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:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

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

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

2 years ago

What is the third ionization energy of titanium?

-Dominant- [34]

Ti^2+(g)-->Ti^3+(g)+-3rd IP=2652.5

5 0

2 years ago

what's the density of an element if a sample has a mass of 43.2 g and a volume of 96.5 mL. (0.45 g/mL)

lisov135 [29]

To find the answer you need to use the formula that will help you to find the density. Density = mass/volume
d = 43.2g/96.5mL = 0.45g/mL

7 0

2 years ago

For an ideal gas, evaluate the volume occupied by 0.3 mole of gas.

Serhud [2]

Answer:

V=0.3×22.4=6.72 liters hope this helps

8 0

2 years ago

In an experiment, mice were fed glucose (C₆H₁₂O₆) containing a small amount of radioactive carbon. The mice were closely monitor

Luden [163]

<h3>Answer:</h3>

Carbon dioxide

Explanation:

Cellular respiration involves a break down of sugars such as glucose to yield energy in the form of ATP and water together with carbon dioxide as byproducts.
In this case, when mice are fed with glucose, cellular respiration takes place breaking down glucose to yield energy in the form of ATP, water and carbon dioxide.
Therefore, the radioactive carbon in glucose showed up in carbon dioxide, since the carbon in glucose ends up in carbon dioxide.

8 0

2 years ago