#1: Which scientist is credited with developing the orbital model of the atom?

A. J.J. Thomson

B. Ernest Rutherford

c. Niels Bohr

D. Democritus
***idk really... I'm guessing jj thomson tho... not too sure.. :/,

The relative atomic mass of copper on a newly discovered planet is [tex]\boxed{63.54{\text{ amu}}}[/tex] .

Further explanation:

An atom is also written as [tex]_{\text{Z}}^{\text{A}}{\text{X}}[/tex] , where A is the atomic mass or mass number, Z is the atomic number, and X is the letter symbol of the element.

Isotopes:

Atoms of the same element with same value of atomic number but different value of mass numbers are called isotopes. Isotopes generally have the same number of protons but the number of neutrons is different.

[tex]_{\mathbf{6}}^{{\mathbf{11}}}{\mathbf{C}}[/tex]  and [tex]_{\mathbf{6}}^{{\mathbf{12}}}{\mathbf{C}}[/tex]  are examples of isotopes. Both are the atoms of the same element carbon. The mass number of [tex]_{\mathbf{6}}^{{\mathbf{11}}}{\mathbf{C}}[/tex]  is 11 while that of  [tex]_{\mathbf{6}}^{{\mathbf{12}}}{\mathbf{C}}[/tex] is 12. But both of these have the same atomic number, i.e, 6.

Relative Atomic Mass:

It is defined as the average mass of an atom of the element compared to 1/12th of the mass of carbon-12 atom. It is represented by [tex]{{\text{A}}_{\text{r}}}[/tex] . The formula to calculate the relative atomic mass of an element is as follows:

[tex]{{\text{A}}_{\text{r}}} = \frac{{\sum {\left\{{\left({{\text{Relative isotopic mass}}}\right)\left({{\text{\%  abundance}}}\right)}\right\}}}}{{100}}[/tex]

Here,

[tex]{{\text{A}}_{\text{r}}}[/tex]  is the relative atomic mass.

It is given that copper (Cu) exists in two isotopic forms, one is 63 Cu, and the other is 65 Cu. So the relative atomic mass of Cu is calculated as follows:

[tex]{{\text{A}}_{\text{r}}}{\text{ of Cu}}=\frac{{\left[\begin{gathered}\left({{\text{Mass of 63 Cu}}}\right)\left({{\text{\%  abundance of 63 Cu}}}\right)+\hfill\\\left({{\text{Mass of 65 Cu}}}\right)\left({{\text{\%  abundance of 65 Cu}}}\right)\hfill\\\end{gathered}\right]}}{{100}}[/tex]                       …… (1)

The mass of 63 Cu is 62.9300 amu.

The percentage abundance of 63 Cu is 69.15 %.

The mass of 65 Cu is 64.9200 amu.

The percentage abundance of 65 Cu is 30.85 %.

Substitute these values in equation (1).

[tex]\begin{aligned}{{\text{A}}_{\text{r}}}{\text{ of Cu}}&=\frac{{\left[{\left( {{\text{62}}{\text{.9300 amu}}}\right)\left({{\text{69}}{\text{.15}}}\right)+\left({{\text{64}}{\text{.9200 amu}}}\right)\left({{\text{30}}{\text{.85}}}\right)}\right]}}{{100}}\\&=\frac{{{\text{4351}}{\text{.6095 amu}}+{\text{2002}}{\text{.782 amu}}}}{{100}}\\&={\text{63}}{\text{.543915 amu}}\\&\approx{\text{63}}{\text{.54 amu}}\\\end{aligned}[/tex]

So the relative atomic mass of copper on a newly discovered planet is 63.54 amu.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603

2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133

Answer details:

Grade: Middle School

Subject: Chemistry

Chapter: Mole Concept  

Keywords: percentage abundance, 63 Cu, 65 Cu, 62.9300 amu, 64.9200 amu, 30.85 %, 69.15 %, 63.54 amu, copper, isotopes, carbon-12, Ar, Cu.

Final answer:

To find the amount of iron that could be obtained from 5.41 grams of a compound that is 69.9% iron, multiply the mass of the compound by the percentage of iron (in decimal form) to get 3.78 grams of iron.

Explanation:

To determine how much iron can be obtained from 5.41 grams of the compound, we use the given percentage composition of iron in the compound, which is 69.9%. We can calculate the mass of iron in the given mass of the compound by multiplying the total mass of the compound by the percentage of iron in decimal form.

To find the mass of iron in 5.41 grams of the compound: (percentage of iron / 100) × total mass of compound = (69.9 / 100) × 5.41 grams = 3.78 grams

Therefore, 3.78 grams of iron could be obtained from 5.41 grams of the compound.

The correct option is A) 1/8

half-life of tritium, or hydrogen-3, is 12.32 years.      

After about 37 years,

= 37/12.32

= 3

means 3 half lives.

sample of tritium that will remain unchanged will be:

= 1/2^no.of half lives

= 1/2³

= 1/2 x 1/2 x 1/2 = 1/8

1/8 remains back

Answer:

-∆G and +Eocell

Explanation:

For an electrochemical reaction to be spontaneous, the change in free energy must be negative. This applies to all chemical reactions as a basic condition for spontaneity of chemical reactions.

More particular to an electrochemical cell is the value of the standard cell voltage. A positive value of standard cell voltage implies a spontaneous electrochemical reaction.

Answer : The correct option is, (C)the number of molecules

Explanation :

When we are comparing 1 mole of oxygen gas [tex](O_2)[/tex] with 1 mole of carbon monoxide gas [tex](CO)[/tex] then we conclude that,

In 1 mole of oxygen gas [tex](O_2)[/tex], there are two number of molecules of oxygen.

In 1 mole of carbon monoxide gas [tex](CO)[/tex], there are two number of molecules (carbon and oxygen).

Hence, the number of molecules must be the same when comparing 1 mole of oxygen gas, with 1 mole of carbon monoxide gas.

The chemical equation representing the reaction of silver nitrate with barium chloride:

[tex] 2AgNO_{3}(aq) + BaCl_{2}(aq)--> 2AgCl (s) + Ba(NO_{3})_{2}(aq) [/tex]

Given mass of barium chloride = 4.62 g

Moles of [tex] BaCl_{2} = 4.62 g BaCl_{2}*\frac{1 mol BaCl_{2}}{208.23 g BaCl_{2}} = 0.0222 mol BaCl_{2} [/tex]

Moles of AgCl = [tex] 0.0222 mol BaCl_{2} * \frac{2 mol AgCl}{1 mol BaCl_{2}} [/tex] = 0.0444 mol AgCl

Mass of AgCl = [tex] 0.0444 mol AgCl * \frac{143.32 g AgCl}{1 mol AgCl} = 6.36 g AgCl [/tex]

The correct option is B: 14/35 × 100.

To determine the percent composition by mass of nitrogen in NH₄OH, we need to use the molar mass of each constituent element and the total formula mass:

(Mass of Nitrogen / Molar Mass of NH₄OH) × 100

Using the given values:

(14 g/mol / 35 g/mol) × 100 = 40%

Thus, the correct option is B: 14/35 × 100.

Answer:

a) 5.40 g B : [tex]3.012\times 10^{23}[/tex] atoms

b) 0.250 mol K : [tex]1.51\times 10^{23}[/tex] atoms

c) 0.0384 mol K : [tex]0.23\times 10^{23}[/tex] atoms

d) 0.02550 g Pt: [tex]7.8\times 10^{19}[/tex] atoms

e)  [tex]1.00\times 10^-{10} g[/tex] Au: [tex]0.03\times 10^{13}[/tex] atoms

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]

a) 5.40 g B

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{5.40g}{11g/mol}=0.5moles[/tex]

1 mole of boron contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus 0.5 moles  of boron contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.5=3.012\times 10^{23}[/tex] atoms

b) 0.250 mol K

1 mole of potassium (K) contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus 0.250 moles of potassium contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.250=1.51\times 10^{23}[/tex] atoms

c) 0.0384 mol K

1 mole of potassium (K) contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus 0.0384 moles of potassium (K) contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.0384=0.23\times 10^{23}[/tex] atoms

d) 0.02550 g Pt

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{0.02550 g}{195g/mol}=1.3\times 10^{-4}moles[/tex]

1 mole of platinum contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus [tex]1.3\times 10^{-4}moles[/tex] of platinum contain=[tex]\frac{6.023\times 10^{23}}{1}\times 1.3\times 10^{-4}=7.8\times 10^{19}[/tex] atoms

e) [tex]1.00\times 10^-{10}g[/tex] Au,

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{1.00\times 10^{-10}g}{197g/mol}=0.005\times 10^{-10}moles[/tex]

1 mole of gold contains = [tex]6.023\times 10^{23}[/tex] atoms

Thus [tex]0.005\times 10^{-10}moles[/tex] of platinum contain=[tex]\frac{6.023\times 10^{23}}{1}\times 0.005\times 10^{-10}=0.03\times 10^{13}[/tex] atoms

An increase in dissolved CO₂ in ocean water results in an increase in acidity (decrease in pH) of the water. This is due to the formation of carbonic acid that dissociates into hydrogen ions. This process has adverse effects on marine life, particularly animals such as corals and shellfish that rely on calcification to build their shells and skeletons.

When carbon dioxide (CO₂) dissolves in the ocean water, it reacts with water to form a weak acid called carbonic acid. This acid then dissociates to form hydrogen ions (H+) and bicarbonate ions (HCO3-). Increased concentration of these hydrogen ions decreases the pH of the water, thus increasing its acidity.

Ocean acidification is detrimental to many marine organisms, especially corals and shellfish. This is because more acidic water interferes with the process of calcification, which they use to build their calcium carbonate shells and skeletons.

The increasing levels of atmospheric CO₂ and rising ocean temperatures are contributing to the acidity of ocean waters, causing harm to marine animals and ecosystems in the process.

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Given:

C= 58.8% ,

H= 9.8% ,

and O=31.4 %

molar mass is = 102 g/mol.

empirical formula=

58.8/C(12.01)= 4.9

9.8/H(1.01)= 9.7

31.4/O(15.99)= 1.96

divide it by the least value so it is 1.96 which shows,

4.9/1.96= 2.5

9.7/1.96= 5

(C2.5H₅O)₂= C₅H₁₀O₂

molecular formula= C₁₀H₂₀O₄

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

Assuming that there is the same amount  of solution X for each reactant

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.

To identify the limiting reagent between sodium hydroxide and crystal violet, normally one would need the reaction equation and stoichiometry. Assuming a hypothetical 1:1 reaction, crystal violet would be the limiting reagent due to its much lower concentration compared to NaOH.

The question asks to identify the limiting reagent in a reaction between sodium hydroxide and crystal violet and calculate the amount of excess reagent remaining after the reaction is complete. Since the concentration of sodium hydroxide (NaOH) is 1.0 M and the concentration of crystal violet is 1.00E-5 M, we do not have the actual reaction equation or the stoichiometry to determine the limiting reagent directly. Normally, the limiting reagent is the one that will be completely consumed first during the reaction, based on the stoichiometry of the reaction. Typically, the number of moles of each reagent would be considered; the one with the fewer moles, per the reaction stoichiometry, would be the limiting reagent.

As an example, if the reaction were 1:1, then clearly crystal violet would be the limiting reagent due to its much lower concentration. To find out how much NaOH remains, we would subtract the moles of NaOH that reacted with crystal violet from the initial moles of NaOH. However, in absence of the reaction equation and assuming a hypothetical 1:1 reaction, all of the crystal violet would react, leaving behind an excess of NaOH. If we had information on the volumes used, we could calculate the actual amount of NaOH remaining using dimensional analysis as demonstrated by various examples provided.

The acidity of the given compounds is determined by how readily they donate H+ ions. By considering their respective positions on the periodic table, it can be inferred that HIO3, iodic acid, is the most acidic as it disperses the resultant negative charge over the largest volume.

The acidity of these chemicals is determined by the strength of the acid, which is determined by the degree to which the acid dissociates in water. In this case, we are looking at a series of oxyacids where the central atom is a halogen: HBrO3 (bromic acid), HClO3 (chloric acid), HFO3 (fluoric acid/supposed to be HF which is hydrofluoric acid), and HIO3 (iodic acid).

Bromic acid, HBrO3, can donate a BrO3- ion and 1 H+ ion, which makes it an acid. Similarly, Chloric acid, HClO3, donates a ClO3- ion and 1 H+ ion. Hydrofluoric acid, HF can donate F- and H+ ions, and Iodic acid, HIO3, donates IO3- and H+ ions.

By noting the position of the halogens on the periodic table, we see that iodine is further down the halogen group than the others. As you go down the group in the periodic table, the atomic radius increases. As such, the resulting negative charge on the larger ions is dispersed over a larger volume, and the acidic strength increases. Because of these, HIO3, iodic acid, is the most acidic of the acids provided.

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The heat of reaction for the combustion of a mole of titanium in the given calorimeter setup is approximately 14765.89 kJ/mol, as determined by calculating the total heat absorbed using the calorimeter's heat capacity and the mass and molar mass of titanium.

When 1.640 g of titanium is combusted in a bomb calorimeter, the resulting temperature increase from 25.00 °C to 76.50 °C can be used to calculate the heat of reaction for the combustion of a mole of titanium. Given that the heat capacity of the calorimeter is 9.84 kJ/°C, we can determine the total heat absorbed by the calorimeter during the reaction.

First, we calculate the total heat absorbed (q) by multiplying the heat capacity (C) by the change in temperature (ΔT):
q = C × ΔT
q = 9.84 kJ/°C × (76.50 °C - 25.00 °C)
q = 9.84 kJ/°C × 51.50 °C
q = 506.34 kJ

Next, to find the heat of reaction per mole of titanium, we convert the mass of titanium to moles using its molar mass (47.87 g/mol):
Number of moles = mass / molar mass
Number of moles = 1.640 g / 47.87 g/mol
Number of moles ≈ 0.0343 mol

Now, we can determine the heat of reaction per mole:
Heat of reaction per mole = q / Number of moles
Heat of reaction per mole ≈ 506.34 kJ / 0.0343 mol
Heat of reaction per mole ≈ 14765.89 kJ/mol

The heat of reaction for the combustion of a mole of Ti in this calorimeter is approximately 14765.89 kJ/mol.

Answer:

∆hf0 of products minus ∆hf0 of reactants

Explanation:

The empirical formula for the compound decomposed in the laboratory which resulted in 1.78 g of Nitrogen and 4.05 g of Oxygen can be calculated by finding the moles of each element and comparing ratios. Based on the calculation, the empirical formula for the compound is NO2.

First, we determine the number of moles of each element involved in the compound. Recall the atomic weight of Nitrogen (N) is about 14 g/mol and Oxygen (O) is about 16 g/mol.

The given weight of Nitrogen is 1.78 g. So, divide this by the atomic weight to get the number of moles: 1.78 g N / 14 g/mol = 0.127 moles of Nitrogen. Similarly for Oxygen, we have 4.05 g so: 4.05 g O / 16 g/mol = 0.253 moles of Oxygen.

To find the empirical formula, we divide each of these by the smallest number of moles, which is for Nitrogen: 0.127/0.127 = 1 for Nitrogen, and 0.253/0.127 = 2 for Oxygen. Hence, the empirical formula for the compound is N1O2 or simply NO2.

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The empirical formula of the compound is N1O2.

To determine the empirical formula of the compound, we need to find the ratio of nitrogen to oxygen. The given data states that 1.78 g of nitrogen and 4.05 g of oxygen were produced. First, convert the mass of each element to moles using their molar masses. The molar mass of nitrogen is 14 g/mol, so 1.78 g of nitrogen is equal to 0.127 moles. The molar mass of oxygen is 16 g/mol, so 4.05 g of oxygen is equal to 0.253 moles. Next, divide each mole value by the smallest mole value to get the mole ratio. In this case, the ratio is approximately 1:2. Therefore, the empirical formula of the compound is N1O2.

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Increase along the Gulf Coast and climate change at sea level. Warming ocean waters expand as temperatures rise, while mountain glaciers and inland ice melt.

Climate is a region's averaged long-term weather pattern over a period of time, usually 30 years. More precisely, it is the average and variation of climatic variables over a period of time that can range from a few months to many millions of years.

The average of the weather is the climate. As an illustration, you can anticipate snow in the Northeast in January or hot and muggy weather in the Southeast in July. Climate is this.

Nearly every element of our life is impacted by the climate, including our food supplies, transportation systems, clothing choices, and vacation destinations. It significantly affects our future, our health, and our means of subsistence. The long-term pattern of the weather in any given location is called the climate.

Thus, increase along the Gulf Coast and climate change at sea level.

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The Gulf Coast and similar regions are experiencing land loss due to sea level rise resulting from melting glaciers and ice sheets.

The loss of land area on the Gulf Coast as oceans expand is primarily a result of sea level rise and coastal erosion. As glaciers and ice sheets melt, including those in Greenland and West Antarctica, sea levels rise at an approximate rate of 3 millimeters per year, contributing to land loss along coastlines. The problem is exacerbated along the Atlantic coast by factors such as low elevations and subsiding land. The coastline from Florida to New York, for example, lost more land than it gained between 1996 and 2011, indicating a negative net change in land area.

Moreover, global assessments reveal that roughly 70% of the world's sandy shorelines are retreating, with a significant proportion of beaches eroding at rates that exceed 0.5 meters per year. Specifically, for every 361 gigatonnes per year of ice mass melting into the ocean, sea levels rise by approximately 1 millimeter. Glaciologists project that if current trends continue, we could see more than 1 meter of sea level rise by 2100, considering both melting land ice and thermal expansion of the ocean water.