A structure that can lessen erosion is

Answer:

1) If life exists in other planets

2) What type of living organisms exists on other planets

3) Predominant atoms/mixture/compound that exists on other planets

Explanation:

Earth and space science is the study of the earth, the space and other planets within the universe.

The following questions can be answered by studying earth and space science

1) If life exists in other planets

Over the years, there has been several reports suggesting this but a deep inquiry into this with scientific data can assist in putting the matter to rest.

2) What type of living organisms exists on other planets

If there is/are lives on other planets, what type of life/living organisms exist there. Are they plants? Are they animals? Or do we have a new classification for them?

3) Predominant atom/mixture/compound that exists on other planets

What is the most popular substance in each of these planets? The role of this substance in the survival of the planet or it's inhabitants also needs to be studied

Answer:

7. Why is the d-block one energy level behind the s-block? Also, why would the f-block be so far behind?

me to

Explanation:

Answer:

Yes, they are balanced.

Explanation:

There are 1 Carbon, 4 Hydrogen and 4 Oxygen on both reactant and product sides.

Answer:

heat production

Explanation:

Answer:

A

Explanation:

Homeostatic imbalance

An example of heterogeneous equilibrium is:

Heterogeneous equilibrium refers to an equilibrium state in a chemical reaction where the reactants and products exist in different physical states or phases. It occurs when substances in different phases, such as solids, liquids, and gases, are involved in a chemical reaction.

Considering the given equations:

The equation I: FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g) represents a heterogeneous equilibrium.

This is because the reactants and products involve different phases (solid and gas). FeO is a solid (s), CO is a gas (g), Fe is a solid (s), and CO₂ is a gas (g). The reaction involves the conversion of a solid and a gas to another solid and a gas, and the equilibrium is established between these different phases.

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The specific heat constant for oxygen is 0.919 J/g K.

\(Q=(10.8)(0.919)(35.2-22.1)\), which is about 130 J (to 3 sf)

The enthalpy change of the reaction C(diamond) → C(graphite) is -2.9 kJ.

The given information is ΔHrxn for the reaction C(diamond) → C(graphite) can be calculated with the given equations:Equations:  C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ  2 CO2(g) → 2 CO(g) + O2(g) ΔH = 566.0 kJ  C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ  2 CO(g) → C(graphite) + CO2(g) ΔH = −172.5 kJThe required reaction can be obtained by adding the equations (1) and (4), as follows:C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)Addition of the two equations (1) and (4) results in a reaction whose products are C(graphite) and CO2.

To get the final equation that involves only the required reactants and products, the equation (2) should be added, which consumes CO2 and produces O2, as shown below:C(diamond) + O2(g) → CO2(g)  ΔH = −395.4 kJ [eq. (1)]  2 CO(g) → C(graphite) + CO2(g)  ΔH = −172.5 kJ [eq. (4)]  2 CO2(g) → 2 CO(g) + O2(g)  ΔH = 566.0 kJ [eq. (2)]  C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)  ΔHrxn=ΣΔHf(products)−ΣΔHf(reactants)  ΔHrxn=[(3 mol CO2)(-393.5 kJ/mol) + (1 mol C(graphite))(0 kJ/mol)] − [(1 mol C(diamond))(0 kJ/mol) + (1 mol O2)(0 kJ/mol) + (2 mol CO(g))(−172.5 kJ/mol)] − [(2 mol CO2)(566.0 kJ/mol)]  ΔHrxn=−2.9 kJ.

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Answer:

D

Explanation:

compound 1 probably ether and compound 3 1-bromo ethane have 2 carbon atoms

There are 2.64 × 1022 molecules of sucrose present in this serving of fruit dessert containing 15.0 g of sugar.

To determine the number of molecules of sugar present in the serving, we need to calculate the number of moles of sugar and then convert it to the number of molecules.

Given:

Mass of sugar (sucrose) = 15.0 g

Molar mass of sucrose (C12H22O11) = 342 g/mol

First, calculate the number of moles of sugar using the formula:

Number of moles = Mass of substance / Molar mass

Number of moles of sugar = 15.0 g / 342 g/mol ≈ 0.0439 mol

Next, we use Avogadro's number, which states that there are approximately 6.022 × 10^23 molecules in one mole of a substance. Therefore, to find the number of molecules of sugar:

Number of molecules = Number of moles × Avogadro's number

Number of molecules of sugar = 0.0439 mol × 6.022 × 10^23 molecules/mol ≈ 2.64 × 10^22 molecules

Therefore, there are approximately 2.64 × 10^22 molecules of sugar present in this serving.

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Answer:

HCl is not a catalyst because these are not used up during the chemical reactions.

Explanation:

Hello there!

In this case, according to the performed experiments, it is possible for us to realize that HCl cannot be a catalyst for this reaction because it is used up during the reaction. This is explained by the fact that catalyst are able to return to the original form once the reaction has gone to completion; this is the example of palladium in the hydrogenation or dehydrogenation of hydrocarbons depending on the case. Moreover, we know that the catalysts increase the reaction rate because they decrease the activation energy of the reaction and therefore the student observed such increase.

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Answer:

Figure 2 represents a nonpolar molecule containing polar covalent bonds

Explanation:

The CO2 molecule is linear, formed by 3 atoms and the electronegativity variations are equal, so its dipolar moment will be null, and this molecule will be nonpolar, however, it has polar covalent bonds.

Answer:

B. Strong, brittle and good insulators.

Explanation:

The correct answer is B. Ceramics tend to be strong, brittle, and good insulators. They are not ductile, chemically stable or malleable.

Answer:

D

Explanation:

Answer: The stoichiometric coefficient for oxygen is \(\frac{1}{2}\).

Explanation:

A number present on the front of an atom, ion or molecule in a chemical reaction equation is called a stoichiometric coefficient.

For example, \(H_{2}O(l) \rightarrow H_{2}(g) + \frac{1}{2}O_{2}(g)\)

Here, the stoichiometric coefficient for \(H_{2}O(l)\) is 1, for \(H_{2}(g)\) is 1 and for \(O_{2}\) is \(\frac{1}{2}\).

Thus, we can conclude that the stoichiometric coefficient for oxygen is \(\frac{1}{2}\).

Answer:

The answer is

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

From the question

volume of benzene = 250 mL

density = 0.8786 g/mL

The mass is

mass = 250 × 0.8786

We have the final answer as

Hope this helps you

The arrow in a chemical reaction can be translated as the following:

A. yields

C. react to form

Explanation:

\(FeSO _{4} + \: KOH \: →K _{2}SO _{4} \: + \: Fe(OH) _{2}\)

The body's muscles become increasingly paralyzed as a result of tetrodotoxin's interference with the signaling process from nerves to muscles. Toxic exposure to tetrodotoxin may be lethal.

Scrub all right, had to drink a poisonous substances that paralyzes body resulting in death.

Tetrodotoxin (TTX), a neurotoxin, is extremely poisonous.

Tetrodotoxin is a sodium channel blocker. It stops neurons from firing action potentials by clinging to the voltage-gated sodium channels in nerve cell membranes and obstructing sodium ions from entering the neuron, which are crucial for an action potential's rising phase.

Due to the neurological system's inability to communicate, muscles are unable to contract in response to nerve inputs. It works by selectively blocking sodium channels.

The majority of its occurrences are in the livers and sex organs of several fish species, such as puffer fish, globefish, and toadfish (order Tetraodontiformes) (gonads).

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Answer:

you have a 3.17molal solution: This is 3.17mol CaCl2 dissolved in 1 litre of water. Make this solution.  

Molar mass CaCl2 = 110.9848 g/mol

3.17mol = 3.17*110.9848 = 351.822g  

Total mass = 1000g H2O + 351.822g CaCl2 = 1,351.822g  

Volume of this solution:

Volume = mass / density

Volume = 1,351.822/1.24  

Volume = 1,090.2 mL  

You have 3.17mol CaCl2 dissolved in 1,090.2 mL solution

Mol CaCl2 dissolved in 1000mL solution =  1000/1090.2*3.17 = 2.91 mol CaCl2  

Molarity of CaCl2 solution = 2.91M

Explanation:

The answer is  21.5 L ,So, start with the balanced chemical equation for the decomposition of hydrochloric acid.

An liquid solution of hydrogen chloride is what is known as hydrochloric acid, also referred to as muriatic acid. It has an unmistakably strong odour and is translucent. It is categorised as a powerful acid. In the digestive processes of the majority of mammal species, including humans, it is a part of the gastric acid. A crucial commercial substance and reagent for laboratories is hydrochloric acid.

Chemical burns from hydrochloric acid can be extremely painful if they come into touch with skin or other organs. Ingestion of hydrochloric acid can result in blindness. The concentration of the acid and how long it remains in touch with the organs determine how severe the burns are.

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Answer:

During the early Paleozoic Era, the Earth's surface was very different from what it is today. The continents were arranged differently, forming one large supercontinent called Pangea. This landmass was surrounded by a single large ocean called Panthalassa. The climate was much warmer and wetter than it is today, with no ice caps at the poles.

The first land plants, known as bryophytes, appeared during the early Silurian Period, around 430 million years ago. These plants were small and simple, lacking roots and vascular tissue. They grew in damp environments, such as along the edges of lakes and streams. They were important in the development of soils and in the colonization of land by other organisms, such as insects and other arthropods.

The rate law and the value of k for the given reaction is 1.7×103 M⁻²s⁻¹. Therefore, option D is correct.

The word "rate law" refers to an expression that expresses reaction rate as the product of the stoichiometric coefficient of the reacting species in a balanced chemical equation multiplied by the molar concentration of the reactants, with each term raised to a power.

\(\rm Rate = k \times [NO]^{n} x [O_{2}]^ {m}\)

Thus, m must = 1

\(\rm Rate\ 1 = k \times [NO 1]^ {m} x [O_{2} 1]^ {n}\\Rate\ 2 = k \time [NO 2]^ {m} x [O_2 2]^ {n}\)

Rearranging this equation

\(Rate1 / [O_2 1]^{n} = k x [NO1]^ {m}Rate2 / [O_2 2]^{n} = k x [NO2]\)

but  [NO 1] = [NO 2]

Hence,

\(Rate1 / [O_2 1]^ {n}= Rate2 / [O_2 2]^{n}\)

Rearranging and substituting in the values

\(([O_2 2] / [O_2 1])^{n} = Rate2 / Rate 1\)

\(2^{n} = 2\)

So, n = 1

Same from run 1 to 3

[NO] doubled

[O₂] stayed constant

Rate quadrupled

\((2)^{n} = 4\)

n = 2

we know that rate = k x [NO]² x [O₂]

Substitute in any value for [NO], [O₂] and rate and calculate K

k = rate / [NO]² x [O₂]

= (8.55x10⁻³ M / sec) / ((0.030M)² x (0.0055M))

= 1.7×103 M⁻²s⁻¹

Thus, option D is correct.

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Your question is incomplete, most probably complete question is

Determine the rate law and the value of k for the following reaction using the data provided.

2 NO(g) + O2(g) 2 NO2(g)

[NO]i (M) [O2]i (M) Initial Rate (M-1s-1)

0.030 0.0055 8.55 x 10-3

0.030 0.0110 1.71 x 10-2

0.060 0.0055 3.42 x 10-2

A. Rate = 57 M-1s-1[NO][O2]

B. Rate = 3.8 M-1/2s-1[NO][O2]1/2

C. Rate = 3.1×105 M-3s-1[NO]2[O2]2

D. Rate = 1.7×103 M-2s–1[NO]2[O2]

The temperature increase from 20 °C to 46 °C indicates that the reaction between zinc and copper sulfate solution is exothermic, with heat being released into the surroundings.

In the given reaction between zinc and copper sulfate solution, the temperature change can provide insights into the type of heat change occurring during the reaction. Based on the provided information, the initial temperature of the copper sulfate solution was 20 °C, and the final temperature of the mixture after the reaction was 46 °C.

The temperature increase observed in this reaction indicates an exothermic heat change. An exothermic reaction releases heat energy into the surroundings, resulting in a temperature rise. In this case, the reaction between zinc and copper sulfate solution is exothermic because the final temperature is higher than the initial temperature.

During the reaction, zinc displaces copper from copper sulfate to form zinc sulfate and copper metal. This displacement reaction is known as a single displacement or redox reaction. Zinc is more reactive than copper and therefore replaces copper in the compound.

The formation of new chemical bonds during the reaction releases energy in the form of heat. This energy is transferred to the surroundings, leading to an increase in temperature. The heat released is greater than the heat absorbed, resulting in a net increase in temperature.

The exothermic nature of this reaction can be explained by the difference in bond energies between the reactants and products. The breaking of bonds in the reactants requires energy input, while the formation of new bonds in the products releases energy.

In this case, the energy released during the formation of zinc sulfate and copper metal is greater than the energy required to break the bonds in copper sulfate and zinc.

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Nitrous acid, phosphoric acid, sulfurous acid and carbonic acid are the  weak acids. Hydrochloric acid and perchloric acid is a strong acid. This can be Written in the aqueous solution.

Nitrous acid is a weak acid and can be written in the aqueous solution as,

HNO2 (aq.) ----> H+ (aq.) + NO2- (aq.)

This is very unstable and exists only in aqueous solution.

Hydrochloric acid is a strong acid and can be written in the aqueous solution as,

HCl(aq.) ------> H+ (aq.) + Cl -(aq.)

perchloric acid  is a strong acid and can be written as

HClO4 (aq.) ---->   H+ (aq.) + ClO4- (aq.)

Phosphoric acid is a weak acid. It is a triprotic acid that exists as a dense liquid. It is an irritant or corrosive to the skin, eyes. It can be written in the aqueous solution as,

H3PO4 (aq.) ----> H+ (aq.) + H2PO4- (aq.)

sulfurous acid is a weak acid and can be written as,

H2SO3 (aq.) ----> H+ (aq.) + HSO3- (aq.)

carbonic acid is a weak acid. Carbonic acid is a carbon-containing compound which has the chemical formula H2CO3. This can be written in the aqueous solution as,

H2CO3 (aq.)  ----> H+ (aq.) + HCO3- (aq.)

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