Which half-reaction correctly describes an oxidation?

Answer:

A) Rapid chemical reaction where gas is released.

Explanation:

Edge 2020 GOT IT RIGHT!

The process of an antacid tablet reacting in water is a rapid chemical reaction characterized by the release of gas. This occurs when components like calcium carbonate in the antacid neutralize stomach acid, producing calcium chloride, water, and carbon dioxide gas.

When an antacid tablet is placed in water, the best description of the process is a rapid chemical reaction where gas is released. Antacid tablets typically contain Arrhenius bases like Al(OH)3, Mg(OH)2, CaCO3, and NaHCO3 that react with and neutralize stomach acid. For example, calcium carbonate reacts with hydrochloric acid in the stomach to produce calcium chloride, water, and carbon dioxide gas, leading to a reduction in acid and sometimes a belch from the released CO2.

CaCO3(s) + 2HCl(aq) = CaCl2(aq) + H2O(l) + CO2(g)

This reaction is an example of a neutralization reaction, which is a type of acid-base reaction, and in the case of antacids, it typically happens rapidly upon the dissolution and reaction of the antacid in the stomach's hydrochloric acid.

Answer:

She added the ingredients in the wrong order, this will make the ingredients to mix in a different way and that can lead to different errors, formation of clumps is one of them.

Answer:

Explanation:

The acid present in the lemon will react with the protein from the milk so it will clabben it forming this little clumps. It could be also the Corn Starch that will not dissolve correctly if add all in a go, due to the polarity.

Answer:

63.5 amu

Explanation:

Answer via Educere/ Founder's Education

The relative atomic mass (r.a.m.) of copper (Cu) is 63.5 atomic mass units (amu).

In the element copper (Cu), the Relative Atomic Mass, often abbreviated as r.a.m, is a value obtained by comparing a given atom's mass to the atomic mass of the isotope carbon-12. Therefore, the r.a.m. of copper (Cu) is 63.5 amu (atomic mass units). This value, depending on the nature of the atoms and molecules involved, may vary slightly. Nevertheless, in most general context, you can use 63.5 amu as the r.a.m. of copper.

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Answer:

5400 cans

Explanation:

First we convert the total weight, 1 ton, to grams:

[tex]1ton=1x10^{6}g[/tex]

Now we need to know the mass of aluminum:

[tex]m_{Al}=\frac{10^{6}*8.1}{100} =81000g[/tex]

Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:

[tex]n=\frac{81000g}{15g/can} =5400cans[/tex]

5,400 cans could be made from 1 ton of the Earth’s crust of aluminum.

We can calculate the number of cans by dividing the total mass of aluminum in earth's crust to the mass of aluminum in one can.

We know that relation between tons and gram is as follow:

1 ton = 10⁶ grams

Given % mass of aluminum = 8.1%

Now, the mass of aluminum in 1 ton can be calculated as:

mass = 8.1 × 10⁶ / 100 = 81,000 grams

Given mass of aluminum in one can = 15 grams

Therefore number of cans = 81,000 / 15 = 5,400

Hence, 5,400 cans can be made.

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Answer:

The answer to your question is:

Explanation:

2.45 g of Phosphorus

MW P = 31 g

MW P2O5 = 2(31) + 5(16) = 142 g

From the balance reaction

               4 P      ⇒ 2 P2O5

Then     4(31) g P    ⇒  2 (142)  g P2O5

       124g  of P       ⇒  284 g   of P2O5              Rule of three

            2.45g P      ⇒      x

x = 2.45 x 284/124 = 695.8/124 = 5.61 g of P2O5

Answer:

5.619  grams of diphosphorus pentoxide are formed when 2.45 g of phosphorus reacts with excess oxygen.

Explanation:

[tex]4P(s)+5O-2(g)\rightarrow 2P_2O_5(s)[/tex]

Mass of  phosphorus = 2.45 g

Moles of phosphorous = [tex]\frac{2.45 g}{31 g/mol}=0.7903 mol[/tex]

According to reaction 4 moles of phosphorus gives 2 moles of diphosphorus pentoxide.

Then 0.7903  moles of phosphorus will give:

[tex]\frac{2}{4}\times 0.7903 mol=0.03957 mol[/tex] of diphosphorus pentoxide

Mass of 0.03957 moles of  diphosphorus pentoxide :

[tex]0.03957 mol\times 142 = 5.619 g[/tex]

5.619  grams of diphosphorus pentoxide are formed when 2.45 g of phosphorus reacts with excess oxygen.

Answer:

Explanation:

Increase of boiling point, decrease of freezing point, and decrease of vapor pressure are all colligative properties.

The colligative properties are the physical properties of the solutions that depends on the number of particles of solute dissolved and not on the nature of the particles.

Regarding the vapor pressure of a solution, since the solute particles occupy part of the surface of the solution, less solvent particles are in the surface and so the vapor pressure decreases. The higher the number of solute particles, the greater the decrease on the vapor pressure.

As for the boiling boiling point, take into account that the boiling points is the temperature at which the vapor pressure of the substance reachs the atmospheric pressrue. Since, the presence of solute particles diminishes the vapor pressure of the solution, the solutions will need to absorb more energy to reach the atmospheric pressure, meaning that the boiling point will be higher. The higher the number of solute particles, the greater the increase on the boiling point.

As for the freezing point, the solutions freeze at lower temperatures when a solute is added. The higher the number of particles of solute, the lower the freezing point of the solution.

Then, you need to find which out of three 1.5 m (molal) solutions of glucose, sodium sulphate, and ammonium phosphate will produce the greater number of particles.

Glucose is a molecular compound (not ionic), so each molecule of glucose will yield one particle.

Sodium sulphate is Na₂SO₄; a ionic compound that ionizes into 3 ions: 2 Na⁺ and 1 SO₄²⁻.

Ammonium phospahte is (NH₄)₃ PO₄; a ionic compound that ionizes into 4 particles: 3 NH₄⁺ and 1 PO₄³⁻.

Therefore, since every unit of ammonium phosphate yields to more particles than every unit of glucose or sodium sulphate, the 1.5 m solution of ammonium phosphate will have the most impact on the colligative properties.

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Given: For Li

Given mass = 2.50 g

Molar mass of Li  = 6.94 g/mol

Moles of Li  = 2.50 g / 6.94 g/mol = 0.3602 moles

Given: For [tex]N_2[/tex]

Given mass = 2.50 g

Molar mass of [tex]N_2[/tex] = 28.02 g/mol

Moles of [tex]N_2[/tex] = 2.50 g / 28.02 g/mol = 0.08924 moles

According to the given reaction:

[tex]6Li+N_2\rightarrow 2Li_3N[/tex]

6 moles of Li react with 1 mole of [tex]N_2[/tex]

1 mole of Li react with 1/6 mole of [tex]N_2[/tex]

0.3602 mole of Li react with [tex]\frac {1}{6}\times 0.3602[/tex] mole of [tex]N_2[/tex]

Moles of [tex]N_2[/tex] that will react = 0.06 moles

Available moles of [tex]N_2[/tex] = 0.08924 moles

[tex]N_2[/tex] is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of [tex]Li_3N[/tex]

1 mole of Li gives 2/6 mole of [tex]Li_3N[/tex]

0.3602 mole of Li react with [tex]\frac {2}{6}\times 0.3602[/tex] mole of [tex]Li_3N[/tex]

Moles of [tex]Li_3N[/tex] = 0.12

Molar mass of [tex]Li_3N[/tex] = 34.83 g/mol

Mass of [tex]Li_3N[/tex] = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

Theoretical yield = 4.18 g

Answer:

The balanced chemical equation for the reaction is:

6 Li + N2 → 2 Li3N

From the equation, we can see that 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N. We can use this information and the molar masses of the elements to calculate the theoretical yield of Li3N:

Molar mass of Li: 6.94 g/mol

Molar mass of N2: 28.02 g/mol

Molar mass of Li3N: 34.83 g/mol

First, we need to calculate the number of moles of Li in the reaction:

2.50 g Li × (1 mol Li/6.94 g Li) = 0.360 mol Li

Next, we need to calculate the limiting reagent in the reaction. Since we have 0.360 moles of Li, and 1 mole of N2 is needed for every 6 moles of Li, we have:

0.360 mol Li × (1 mol N2/6 mol Li) = 0.0600 mol N2

Therefore, N2 is the limiting reagent in the reaction. We can now calculate the theoretical yield of Li3N:

0.0600 mol N2 × (2 mol Li3N/1 mol N2) × (34.83 g Li3N/1 mol Li3N) = 4.74 g Li3N

Therefore, the theoretical yield of Li3N is 4.74 g.

Answer:

A. 97.8 mm  

Explanation:

There are 10 divisions between 9 cm (90 mm) and the end of the ruler (10 cm or 100 mm).

Each division equals 1 mm.

Each 5 mm mark has a longer tick, and the dashed red line is between the 97 mm and 98 mm marks.

You would normally estimate to the nearest tenth of a division. An estimate of 0.8 mm is reasonable.

The length of the object is 97.0 mm + 0.8 mm = 97.8 mm.

B is wrong. You can't possibly estimate to the nearest hundredth of a division.  

C is wrong. The dashed red line slightly before the 98 mm mark.

D is wrong. If the dashed red line were exactly on the 98 mm mark, you would record the measurement as 98.0 mm. This indicates that you measured the object to zero tenths on either side of the mark.

Answer:

1propyl,3Ethyl,3methyl pentane

Explanation:

An enzyme is defined as a substance which acts as a catalyst and it is produced by a living organism to bring about a specific biochemical reaction.

An enzyme actually helps in increasing the rate of a reaction by combining at a specific site and like a catalyst an enzyme does not get consumed in a chemical reaction.

An enzyme binds with the substrate and results in the formation of a stable enzyme-substrate complex. Hence, [tex]\Delta G[/tex] will be lower when compared to both uncatalyzed and loosely bound reaction.

An enzyme also helps in lowering the activation energy so that more number of molecules can participate in the reaction. Hence, it leads to increase in the rate of reaction.

Thus, we can conclude that qualities of a chemical reaction which are affected by enzymes are as follows.

The quality of a chemical reaction that is affected by enzymes would be the energy required to start the reaction.

The energy required to start a reaction is known as the activation energy. It is a barrier that the reactants must cross before they can become products.

Enzymes do not interfere with the speed of reaction directly. Instead, they lower the energy barrier, the activation energy, that the reactants must cross and as such, products are arrived at quicker.

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a) mass = 4.69gms b) mass = 5.5gms c) mass = 5.15gms d) mass = 5.94 gms.

Part A: Ba(s)+Cl₂(g)⇄BaCl₂(s)

Ba= 137

so,moles of Ba= 3.09g /137=0.02255 moles

so that the moles of BaCl₂

so mass= 0.02255x(137 +35.5x2)=4.69gms

Part B: CaO(s)+CO₂(g)⇄CaCO₃(s)

molar mass of Cao=40+16=56

moles=3.09/56=0.055

so ,moles of CaCO₃= 0.055

so mass= 0.055x(100)=5.5gms

Part C: 2Mg(s)+O₂(g)⇄2MgO(s)

Mg= 24 gms per mole

so ,no. of moles= 3.09 /24=0.128

so same no. of moles for MgO= 0,.128

weight=0.128x40=5.15gms

Part D: 4Al(s)+3O₂(g)⇄2Al₂O₃(s)

the molar mass of Al= 26

so,no.of moles = 3.09/26=0.1188

4moles of Al gives 2 moles of Al₂O₃

so, 0.1188 will give 0.0594 moles of Al₂O₃

weight= (26x2 +16x3) x0.0594=5.94 gms

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In the chemical reactions given, the mass of the product formed from 3.68g of the underlined reactant varies based on stoichiometry ranging from 2.98g to 3.68g for different reactions.

The subject question is related to the calculation of the mass of products in a chemical reaction using stoichiometry. Here is how you can calculate the mass for each reaction:

Part A: Ba(s)+Cl₂(g) → BaCl₂(s)
Since one mole of Ba produces one mole of BaCl2, we calculate the molar mass of BaCl2 (208.23 g/mol) and find that 3.68g of Ba will yield 3.68g of BaCl2.

Part B: CaO(s)+CO₂(g) → CaCO₃(s)
In this reaction, one mole of CaO reacts to form one mole of CaCO3. We know the molar mass of CaCo3 is 100.1 g/mol. Using stoichiometry, we calculate the mass to be 3.68g of CaCO3.

Part C: 2Mg(s)+O₂(g) → 2MgO(s)
Here, 2 moles of Mg yield 2 moles of MgO. Given that the molar mass of MgO is 40.3 g/mol, we find that 3.68g of Mg will yield 3.68g of MgO.

Part D: 4Al(s)+3O₂(g) → 2Al₂O₃(s)
In this case, 4 moles of Al yields 2 moles of Al2O3. Considering the molar ratio, we find that 3.68g of Al will produce 2.98 g of Al2O3.

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1) Answer:

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2) Answer:

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3) Answer:

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4) Answer:

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5) Answer:

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6) Answer:

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7) Answer:

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

Final answer:

The questions involve applications of the ideal gas law and concepts of molar volume at standard temperature and pressure (STP). Answers were calculated using ideal gas law formulas, including conversions of conditions to find moles of gases and volumes at STP.

Explanation:

To answer these questions, let's apply the ideal gas law PV = nRT, where P is pressure in atmospheres (atm), V is volume in liters (L), n is the number of moles of gas, R is the ideal gas constant (0.0821 L atm/mol K), and T is temperature in Kelvin (K). Additionally, knowledge of the molar volume of a gas at standard temperature and pressure (STP, defined as 0°C or 273.15 K and 1 atm pressure), which is 22.4 L/mol, is vital for some of the questions.

Question 1

Using the ideal gas law, rearranged to n = PV/RT, with V = 37.5 L, P = 1.25 atm, and T = 307 K, we can calculate the moles of oxygen produced. This calculation yields approximately 1.86 mol O2.

Question 2

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 1.5 moles of oxygen gas would occupy 34 L (1.5 * 22.4 L/mol).

Question 3

First, convert the mass of Zn to moles, then apply the stoichiometry of the reaction to determine moles of H2 produced. Using the ideal gas law under the given conditions, the volume of H2 can be calculated, resulting in approximately 2.29 L H2.

Question 4

O2 produced from 2.80 liters of CO2 at STP would be half, due to the mole ratio of CO2 to O2 in the equation. Therefore, the correct answer is 1.40 liters of O2.

Question 5

Knowing the volume of H2 gas produced and using its molar volume at STP, calculate the moles of H2, and from there, use stoichiometry to determine the mass of Na metal initially used, yielding 10.0 grams of sodium metal.

Question 6

This statement is false; while at STP (273.15 K, 1 atm), one mole of a gas occupies 22.4 L, the conditions specified in the question (298 K, 1 atm) do not define STP.

Question 7

At STP, the volume of one mole of gas is 22.4 liters.

Answer:

The pH is 4.76 (option C)

Explanation:

pH of a solution = TO BE DETERMNED

[sodium acetate] = 1M   ;   [acetic acid] = 1M

K = [H+] [A-] / [HA] 

K = [H+] [1] / [1] 

K = H  

pH = pK

⇒ now pH is equal to pK

⇒ The value of pK is given which is 4.76

⇒ This means the value of the pH is also 4.76

Answer:

The pH is 4.76 (option C)

Explanation:

Explanation:

pH of a solution = TO BE DETERMNED

[sodium acetate] = 1M   ;   [acetic acid] = 1M

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

[tex]P_{B}=x_{B}*P_{B}^{o}[/tex]

[tex]P_{T}=x_{T}*P_{T}^{o}[/tex]

Where:

[tex]P_{B}[/tex] is partial pressure for benzene in the liquid  

[tex]x_{B}[/tex] is benzene molar fraction in the liquid  

[tex] P_{B}^{o}[/tex] vapor pressure for pure benzene.  

The total pressure in the solution is:

[tex]P= P_{T}+ P_{B}[/tex]

And  

[tex]1=x_{B}+x_{T}[/tex]

Working on the equation for total pressure we have:

[tex]P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}[/tex]

Since [tex]x_{T}=1-x_{B}[/tex]

[tex]P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}[/tex]

We know P and both vapor pressures so we can clear [tex]x_{B}[/tex] from the equation.

[tex] x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}[/tex]

[tex] x_{B}=\frac{35- 22}{75-22} = 0.24[/tex]

So  

[tex]x_{T}=1-0.24 = 0.76[/tex]

To get the mole fraction for the vapor we know that in the equilibrium:

[tex]P_{B}=y_{B}*P[/tex]

[tex]y_{T}=1-y_{B}[/tex]

So  

[tex]y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}[/tex]

[tex] y_{B}=\frac{0.24*75}{35}=0.51[/tex]

[tex]y_{T}=1-0.51=0.49[/tex]

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

Answer:

The answer to yourquestion is: a) H₂CNN

Explanation:

Remember that H has one valence electron

                          C has four valence electrons

                          N has five valence electrons

I recommend you draw the Lewis structure to decide which isthe best arrengement. The Lewis structure is in the attachment.

The most plausible atomic arrangement for cyanamide (CH 2 N 2 ) is likely H2NCN, as the formal charges of the atoms in this configuration is closest to their natural state. The formal charge is calculated using an equation that considers valence and bonding electrons. In H2NCN, both nitrogen atoms have a charge of -1 while the carbon remains neutral.

To determine the most plausible atomic arrangement for cyanamide (CH 2 N 2 ), we need to look at their formal charges. For the three arrangements given, H2CNN, H2NCN, and HNNCH, we can calculate the formal charges by using the formula: Formal Charge = [# of valence electrons on atom in free state] - [non-bonding electrons + 1/2(bonding electrons)].

Considering cyanamide, nitrogen typically has a charge of -3, carbon is typically neutral, and hydrogen is typically +1. Therefore, looking at the three arrangements, H2NCN seems to have the most plausible arrangement based on the formal charges. In H2NCN, both nitrogen atoms have a charge of -1 while the carbon remains neutral, which is closest to the natural state of these elements.

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Answer:

5.122 × 10⁻¹³

Explanation:

Sphere volume is calculated with the formula:

[tex]V = \frac{4}{3}[/tex] π [tex]r^{3}[/tex]

V - volume

r - radius

radius = diameter / 2

radius of atom= 1 / 2 = 0.5 Å

radius of nucleus = (8 × 10⁻⁵) / 2 = 4 × 10⁻⁵ Å

now we may calculate the volume for the atom and the nucleus

atom volume = (4/3) × 3.14 × (0.5)³ = 0.523 ų

nucleus volume = (4/3) × 3.14 × (4 × 10⁻⁵)³ = 2.679 × 10⁻¹³ ų

now the fraction of volume of the atom that is taken up by the nucleus:

fraction = nucleus volume / atom volume

fraction = 2.679 × 10⁻¹³ / 0.523 = 5.122 × 10⁻¹³

Answer: The Oxygen will have a partial negative charge, and the hydrogen will have a partial positive charge.

Explanation:

Electronegativity is the tendency of an atom to attract a pair of electrons when forming a chemical bond. The more electronegative atom will attract the electrons more, and will have a partial negative charge, because the electrons are negatively charged. The less electronegative atom will have the electrons the other atom attracted further away from it, so it will have a partial positive charge.

Oxygen is more electronegative than hydrogen. Oxygen has a nuclear charge of 16 protons (positively charged), whereas hydrogen only has 1. As a result, the pull these 16 protons produce on the electrons will be stronger than the pull only 1 proton produces, and the electrons will be closer to the Oxygen atom.

Answer:

mixture

Explanation:

Chromatography separates components of a material based on their difference in polarity. Since the chromatography experiment resulted in more than one mark appearing, there must be more than one component in the material.

Upon standing for 24 hours, the liquid ink separates into its different components based on their differing properties, such as density or solubility. Since a pure substance is homogeneous and has the same properties throughout, the ink cannot be a pure substance.

Answer:

Th answer to your question is:

a)  3.5 x10⁻¹⁰ meters; 0.35 nm

b) 6857142.86 atoms

c) Volume = 2.06 x 10⁻²³ cm³

Explanation:

a) data

Uranium atoms = 3.5A°

meters

           1 A° ----------------  1 x 10 ⁻¹⁰ m

         3.5A° ---------------  x

 x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters

          1 A° ------------------ 0.1 nm

        3.5 A° ---------------- 0.35 nm

b) 2.4 mm

Divide 2,40 mm / uranium diameter

But, first convert 3,5A° to mm   = 3.5 x 10⁻⁷ mm

# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86

c) volume in cubic cm

Convert 3.5A° to cm  = 3.5 x 10⁻⁸

Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³

Volume = 2.06 x 10⁻²³ cm³

a) The radius of a uranium atom is approximately 1.75 Å, which is equivalent to 17.5 nm. b) To span a distance of 2.40 mm, approximately 6.86 billion uranium atoms would be needed. c) The volume of a single uranium atom, assuming it is a sphere, is (4/3) * π * (1.75 [tex]Å)^3.[/tex]

a.) To find the radius of a uranium atom, we can divide the given diameter by 2. The formula to convert the radius from angstroms (Å) to meters (m) is:

(radius in m) = (diameter in m) / 2

Given that the diameter is 3.50 Å, we can calculate the radius:

(radius in m) = (3.50 Å / 2) = 1.75 Å

Next, to convert the radius from angstroms to nanometers (nm), we can use the conversion factor:

(radius in nm) = (radius in Å) * 10

Therefore, the radius of a uranium atom is 1.75 Å (in meters) and 17.5 nm (in nanometers).

b.) To find how many uranium atoms would be needed to span 2.40 mm, we can divide the distance by the diameter of a uranium atom:

Number of uranium atoms = (2.40 mm) / (diameter of a uranium atom)

To calculate this, we need to convert the diameter from angstroms to millimeters using the conversion factor:

1 mm = 10000000 Å

Therefore, the number of uranium atoms needed to span 2.40 mm is:

Number of uranium atoms = (2.40 mm) / (3.50 Å * (1 mm / 10000000 Å))

c.) To find the volume of a single uranium atom, assuming it is a sphere, we can use the formula for the volume of a sphere:

Volume = (4/3) * π * (radius)^3

Using the radius we calculated in part a, we can substitute the value into the formula and solve for the volume:

Volume = (4/3) * π * (1.75[tex]Å)^3[/tex]

Since the volume is given in cubic centimeters (cm³), we need to convert the unit of length from angstroms to centimeters:

1 Å = 1E-8 cm

Therefore, the volume of a single uranium atom is:

Volume = (4/3) * π * (1.75 Å * (1E-8 cm [tex]/ 1 Å))^3[/tex]

Once we calculate this volume, we will have the answer in cubic centimeters (cm³).

Answer:

The correct answer is option D: Reheat the sample until its mass is constant.

Explanation:

The student should reheat the sample until its mass will be constant. This to avoid that the percentage contains other substances than only water. During heating, hydrated salt loses its water. He only wants to know the percentage of water, so until the mass keeps changing a lot ( isn't constant), means there is still both water and salt. If the mass doesn't change anymore, means that all the water has been heated and evaporated. Then he can calculate the percentage of salt in the sample, and thus the percentage water as well.

Answer:

The answer to your question is: Alkyl halide

Explanation:

Ether: is a functional group in which an oxygen is attached to two alkyl or aryl groups.

                   R ---  O --- R'        

Alcohol: is a functional group which has one or more hydroxyl group in its structure.

                  R ---- OH

Alkyl halide: is a group in which one or two hydrogens are replaced by halogens.

                  R --- X      X = halogen

Carbonyl group: is a structure in which one carbon is double-bounded to one oxygen

                       R --- C = O

                                |

                               R'    

Answer:

Alkyl halide is the right answer

Answer:

The answer to your question is: 63.5 u

Explanation:

Data                  Percent                         Molecular mass

63 Cu               69.2%                             62.9296 u

65Cu                 30.8%                             64.9278 u

Average atomic mass = ?

Formula

Average atomic mass = (0.692 x 62.9296) + (0.308 x 64.9278)

                                     = 43.55 + 19.99

                                      = 63.5 u

The average atomic mass of copper is calculated by multiplying the atomic mass of each isotope (63 Cu and 65 Cu) by its natural abundance, then adding those values together. The result is approximately 63.57 u.

The average atomic mass of a chemical element like copper is calculated by multiplying the atomic mass of each isotope by its natural abundance (as a decimal), then adding those values together. In this case, copper has two common isotopes: 63Cu and 65Cu.

Here's how you'd calculate it:

So, the average atomic mass of copper is approximately 63.57 u.

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The answer is option B "era." A era can be up to several millions of years. It isn't period because period can be days to years, it isn't years because it's only one year, wouldn't be option D either because a epoch is up to three million years.

Hope this helps.

Answer:

The answer to your question is: KOH

Explanation:

Formula

Molarity = # of moles / volume

Process

a. No right answer.

b. 121.45 g of KOH in 75.0 mL

MW KOH = 56 g

                                    56 g --------------------- 1 mol

                                  121.45 g -----------------   x

                                x = (121.45 x 1) / 56

                               x = 2.17 mol

M = 2.17 / 0.075

M = 29

c. 23.49 g of NH4OH in 125.0 mL

MW NH4OH = 35 g

                                   35 g  --------------------  1 mol

                               23.49 g  ---------------------   x

                                   x = (23.49 x 1) / 35 = 0.67 mol

M = 0.67 / 0.125

M = 5.36

d. 217.5 g of LiNO3 in 2.00 L

MW LiNO3 = 69 g

                                69 g ----------------------  1 mol

                              217.5 g ----------------------  x

                                 x = (217.5 x 1) / 69 = 3.15 mol

M = 3.15 / 2

M = 1.6

e. 15.25 g of Pb(C2H3O2)2 in 65.0 mL

MW Pb(C2H3O2) = 266 g

                               266 g ------------------ 1 mol

                                 15.25g ...................   x

                              x = (15.25 x 1) / 266 = 0.06 mol

M = 0.06 / 0.065

M = 0.92

Answer:

The answer to your question is below

Explanation:

Barium sulphate (BaSO₄) is white solid crystal inorganic compound that is odorless and insoluble in water.

Sodium chloride (NaCl) is also a white solid crystal inorganic compound odorless and soluble in water.

The, if we try to separate a mixture of these compounds we can do it by dissolving them in water, NaCl will dissolve and BaSO₄ will precipitate.

Later, we can filtrate the sample to obtain the  BaSO₄, in a filtrate paper, and the NaCl will be obtain by evaporation.

To separate barium sulfate from sodium chloride, add water and stir the mixture. Sodium chloride will dissolve, while barium sulfate remains as a solid. Use filtration to separate the two.

To separate barium sulfate (BaSO4) from sodium chloride (NaCl), you can use the principle of solubility. Since barium sulfate is insoluble while sodium chloride is soluble in water, you can add water to the mixture and stir it. The sodium chloride will dissolve in the water, forming a solution, while the barium sulfate will remain as a solid precipitate. You can then use filtration to separate the solid barium sulfate from the liquid sodium chloride solution.

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Answer:

The answer to your questions are

a) 190.96 gr

b) 0.387 moles

c) 2.332 x 10²³ molecules

d) 10 atoms

Explanation:

a)              192.217 u = (192.963)(0.627) + 0.373x

            0.373x = 192.217 - 120.99

             0.373x = 71.23

                   x = 190.96 g

b) MW C5H10O5 = (12 x 5) + ((10 x 1) + (16 x 5) = 150 g

                            = 60 + 10 + 80

                           = 150 g

                150 g ---------------- 1 mol

                  58 g  -------------    x

x = 58 x 1/150 = 0.387 moles

                   1 mol ---------------- 6 .023 x 10²³ molecules

                0.387 mol -----------   x

 x = 0.387 x 6 .023 x 10²³/1 = 2.331 x 10²³ molecules

Answer:

Na+ flows into the cell and then K+ flows out of the cell

Explanation:

When the cell is at rest potential, Na+ is more highly concentrated on the outside of the cell, and K+, on the inside of the cell. During an action potential, the plasma membrane depolarizes, Na+ channels are opened and Na+ starts flowing into the cell, in favor of its electrochemical gradient. When this happens, K+ will also be opened and K+ will start flowing outside the cell, causing the membrane potential to go back to its rest value.

Answer : The molecular formula of a compound is, [tex]CHCl_3[/tex]

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 0.1006 g

Mass of H = 0.0084 g

Mass of Cl = 0.8910 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.1006g}{12g/mole}=0.0084moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0084g}{1g/mole}=0.0084moles[/tex]

Moles of Cl = [tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{0.8910g}{35.5g/mole}=0.0251moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{0.0084}{0.0084}=1[/tex]

For H = [tex]\frac{0.0084}{0.0084}=1[/tex]

For Cl = [tex]\frac{0.0251}{0.0084}=2.98\approx 3[/tex]

The ratio of C : H : Cl = 1 : 1 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_1H_1Cl_3=CHCl_3[/tex]

The empirical formula weight = 1(12) + 1(1) + 3(35.5) = 119.5 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]

[tex]n=\frac{119.6}{119.5}=1[/tex]

Molecular formula = [tex](CHCl_3)_n=(CHCl_3)_1=CHCl_3[/tex]

Therefore, the molecular of the compound is, [tex]CHCl_3[/tex]

The molecular formula of the compound is C4H2Cl4 (carbon tetrachloride).

To find the molecular formula of the compound, we need to determine the empirical formula first. The empirical formula represents the simplest whole number ratio of the elements present in a compound. We can calculate the empirical formula by converting the percentages of the elements to moles and finding the mole ratios.

In this case, the compound contains 10.06% carbon, 89.10% chlorine, and 0.84% hydrogen. Assuming a 100g sample, we have 10.06g of carbon, 89.10g of chlorine, and 0.84g of hydrogen. Converting these masses to moles, we find that the mole ratio is approximately 0.835 moles of carbon, 2.517 moles of chlorine, and 0.835 moles of hydrogen.

The molecular formula can then be determined by comparing the experimental molar mass (119.6 g/mol) with the empirical formula mass. Given that the empirical formula mass is 2*(12.01g/mol) + 35.45g/mol + 2*(1.01g/mol) = 62.02 g/mol, we can divide the experimental molar mass by the empirical formula mass to find the whole number ratio. The ratio is approximately 1.93, which can be rounded to 2.

Therefore, the molecular formula of the compound is C4H2Cl4 (carbon tetrachloride).

Answer:

D

Explanation:

Ketone is R-CO-R' Which is D

A is R-CHO

B is R-COO-R'

C is R-COOH

Answer : The correct option is, (D)

Explanation :

A ketone is an organic compound which contain a functional group with (-CO) i.e R-CO-R'

R-CO-R' functional group consists of a carbonyl center that means a carbon which is double-bonded to the oxygen. The carbon of this carbonyl group bonded with a carbon chain, (R) to one end and a carbon chain, (R') to another end.

For example : [tex]H_3C-CO-CH_3[/tex]

In the given options we conclude that, option D is a ketone.

Hence, the correct option is, (D)