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Anna11 [10]
3 years ago
9

What forms when all of the continents on earths surface merge into one massive landform?

Chemistry
1 answer:
Nutka1998 [239]3 years ago
6 0
Continents are generally the largest landforms on Earth. The Eurasian continent, a combination of Europe and Asia (since they are in contact for thousands of miles along the Ural Mountains) is by far the largest landform. Even without including Europe, Asia is the largest of the individual continents.
Hope this helped.
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If 5 moles of KCIO3 react completely how many moles of KCI will firm
Lubov Fominskaja [6]
With what KClO3 reacts with if it is by temperature
2KClO3===》KCl + KClO + O2
4 0
3 years ago
Which of the following has the smallest first ionization energy: F, Be, Sr or Ra?
topjm [15]
Ra, would have the lowest ionization energy. Remember ionization energy increases going up and to the right.
8 0
3 years ago
How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g
Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

<u>Explanation:</u>

First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

The energy released for 230 grams of decay of thorium is 2.0916\times 10^{-13}J

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

Hence, the energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

5 0
3 years ago
The water-gas shift reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is used industrially to produce hydrogen. The reaction enthalpy is ΔH∘=−4
Pani-rosa [81]

Answer:

The answers are in the explanation

Explanation:

A. For the reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g);  ΔH°=−41kJ.

As the reaction is exothermic ( ΔH°<0), you need to use low temperature to increase the equilibrium yield of hydrogen -LeChatelier's principle-.

We would use <em>low </em>temperature. For an <em>exothermic </em>reaction such as this, <em>decreasing </em>temperature increases the value of K and the amount of products at equilibrium.

B.

c. No. We cannot increase the equilibrium yield of hydrogen by controlling the pressure of this reaction.

It is possible to increase the equilibrium yield of reaction by controlling the amount of reactants added. As reactants and products are gases, the pressure of the reaction will not change the amount of reactants or products in the equilibrium.

I hope it helps!

8 0
3 years ago
A student prepares a aqueous solution of trimethylacetic acid . Calculate the fraction of trimethylacetic acid that is in the di
irinina [24]

The question is incomplete, here is the complete question:

A student prepares a 0.21 mM aqueous solution of trimethylacetic acid. Calculate the fraction of trimethylacetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 3 significant digits.

<u>Answer:</u> The percent of dissociation of trimethylacetic acid is 19.5 %

<u>Explanation:</u>

We are given:

Concentration of trimethylacetic acid = 0.21 mM  = 0.00021 M    (Conversion factor:  1 mole = 1000 millimoles )

The chemical equation for the dissociation of trimethylacetic acid follows:

                     C_4H_9COOH\rightleftharpoons C_4H_9COO^-+H^+

<u>Initial:</u>                   0.00021

<u>At eqllm:</u>             0.00021-x              x            x

The expression of K_a for above equation follows:

K_a=\frac{[C_4H_9COO^-][H^+]}{[C_4H_9COOH]}

We know that:

K_a\text{ for }C_4H_9COOH=1\times 10^{-5}

Putting values in above expression, we get:

1\times 10^{-5}=\frac{x\times x}{(0.00021-x))}\\\\x=0.000041,-0.000051

Neglecting the negative value of 'x' because concentration cannot be negative.

To calculate the fraction of dissociation, we use the equation:

\text{Percent of dissociation}=\frac{[H^+]}{[C_4H_9COOH]}\times 100

Putting values in above equation, we get:

\text{Percent of dissociation}=\frac{0.000041}{0.00021}\times 100\\\\\text{Percent of dissociation of trimethylacetic acid}=19.5%

Hence, the percent of dissociation of trimethylacetic acid is 19.5 %

8 0
3 years ago
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