Which description best matches Rutherford's model of the atom?
Group of answer choices
A The atoms is a positive sphere with negative electrons embedded in it.

B Electrons orbit the nucleus in set, circular orbits.

C The atoms has a tiny, massive nucleus at its center with electrons surrounding it

D The atoms is a solid, indivisible sphere

Answer: The answer is A.

Explanation: Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions. Calcium is also classified chemically as one of the alkaline earth elements (that is, in Group 2 of the periodic table. The metal is obviously reactive.

Answer:

D) Alkaline Earth Metals

Explanation:

:)

Answer:

The number of moles of formic acid needed is 4.5x10⁻³ moles.

Explanation:

We can find the moles of formic acid using the Henderson-Hasselbalch equation:

[tex] pH = pKa + log(\frac{[CHOO^{-}]}{[CHOOH]}) [/tex]

We know:

pH = 3.60

pKa = 3.75                    

By solving the above equation for [CHOO⁻]/[CHOOH] we have:

[tex] \frac{[CHOO^{-}]}{[CHOOH]} = 10^{(pH - pKa)} = 10^{(3.60 - 3.75)} = 0.71 [/tex]

[tex] [CHOO^{-}] = 0.71[CHOOH] [/tex]  (1)

Now, we have:

[tex] [CHOOH] + [CHOO^{-}] = 0.03 M [/tex]   (2)

By entering equation (1) into (2) we have:

[tex] [CHOOH] + 0.71[CHOOH] = 0.03 M [/tex]

[tex] [CHOOH] = 0.018 M [/tex]

Hence, the concentration of formate is:

[tex] [CHOO^{-}] = (0.03 - 0.018)M = 0.012 M [/tex]

Finally, the number of moles of formic acid is:

[tex] n_{CHOOH} = [CHOOH]*V = 0.018 \frac{mol}{L}*0.250 L = 4.5 \cdot 10^{-3} moles [/tex]

Therefore, 4.5x10⁻³ moles of formic acid are needed.

I hope it helps you!  

We have that for the Question "How many moles of formic acid will you need?"

It can be said that

From the question we are told

to prepare 250.0 mL of a pH=3.60, total buffer concentration of formic acid + formate of 0.030 M, The pKa of formic acid is 3.75.

Let [tex]1+COO4 = X, 1+COON = (0.03-X)[/tex]

[tex]pH = pKa + log\frac{1+COON}{1+COO4}\\\\3.6 = 3.75 +log\frac{0.03-X}{X}\\\\log\frac{0.03-X}{X} = 3.6 - 3.75\\\\\frac{0.03-X}{X} = 0.708\\\\X = 0.0175M[/tex]

Therefore,

[tex]moles of formic acid = 0.0175*0.25mole\\\\= 4.39*10^{-3}mole[/tex]

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Answer:

The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH 3CH 2Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names.05‏/06‏/2019

The IUPAC name of this compound is 2,3- dichlorobutane.

Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.

Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:

1)Molecular compounds where in atoms are joined by covalent bonds.

2) ionic compounds where atoms are joined by ionic bond.

3)Inter-metallic compounds where atoms are held by metallic bonds

4) co-ordination complexes where atoms are held by co-ordinate bonds.

They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.

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Answer:

solubility

Explanation:

Explanation:

see the pic for the answer

The energy of a photon that would move an electron from level 2 to level 5 : B. 4.8 eV

The electron energy at the nth shell can be formulated:

Rh = constant 2.179.10⁻¹⁸ J

So the electron transfer energy (ΔE)

ΔE = E final- E initial

energy at n=2(level 2) = -5.3 eV

energy at n=5(level 5) = -0.5 eV

So the energy absorbed :

[tex]\tt \Delta E=-0.5-(-5.3)=-0.5+5.3=4.8`eV[/tex]

The energy required to move an electron from level 2 to level 5 is 4.8 eV.

According to the Bohr model of the atom, an electron moves from a lower to a higher energy level when it absorbs energy. The same energy is released when the atom returns to ground state.

When an electron from level 2 to level 5, the energy of the photon required is;

ΔE = E5 - E2 = -0.5eV - (-5.3eV)

ΔE = 4.8eV

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Answer:

The correct answer is c. Distillation

Explanation:

The volatility of a substance refers to the tendency to vaporize. The more volatility, more readily the substance vaporizes and passes from liquid state to gas state.

From the options, the only operation that separes substances by their tendency to vaporize is distillation. In distillation, evaporation and condensation processes at different temperatures are used to separe miscible liquid substances.

Therefore, the correct option is c. Distillation

Answer:

C. Distillation

Explanation:

EDGE2021

Answer:

6.022×10²² molecules

Explanation:

Given data:

Volume of nitrogen = 224 L

Pressure = standard = 1 atm

Temperature = standard = 273 K

Number of molecules = ?

Solution:

PV = nRT

1 atm × 224 L = n × 0.0821 atm.L/mol.K × 273 K

224 atm.L = n ×22.41 atm.L/mol

n = 224 atm.L/22.41 atm.L/mol

n = 10 mol

1 mole contain 6.022×10²³ molecules

10 mol×6.022×10²³ molecules/ 1 mol

60.22×10²³ molecules

6.022×10²² molecules

Answer:

thank you! ladies... love the answer is A) 6.022 x 10 ^22

Explanation:

love this one!

have a good one everyone

It takes 333.3 s for Oxygen to diffuses

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or  

the effusion rates of two gases = the square root of the inverse of their molar masses:  

[tex]\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }[/tex]

or  

[tex]\rm M_1\times r_1^2=M_2\times r_2^2[/tex]

r₁ H₂ = 300 cm³/50 s=6 cm³/s

M₁ H₂ = 2 g/mol

M₂ O₂ = 32 g/mol

[tex]\tt 2\times 6^2=32\times r_2^2\\\\r_2^2=\dfrac{2\times 6^2}{32}=2.25\rightarrow r_2=1.5[/tex]

the diffusion time of Oxygen :

[tex]\tt r_2=\dfrac{V}{t}\\\\t=\dfrac{V}{r_2}=\dfrac{500~cm^3}{1.5~cm^3/s}=333.3~s[/tex]

Answer:

The pressure of the gas is kept constant. If the Kelvin temperature of the gas is doubled, the volume of the gas is. O 1.

Answer:

1.1

Explanation:

The slope of a line can be calculated using the values of the x and y corrdinates. The equation is given as;

M = ΔY /  ΔX = Y2 - Y1 / X2 - X1

From the points;

(25.6cm³ , 28.16g) - (X1, Y1)

(17.3cm³, 19.03g) - (X2,  Y2)

Inserting the values into the equation;

M = 19.03 - 28.16 /  17.3 - 25.6

M = -9.13 / -8.3 = 1.1

Answer:

D. It is a Mixture

Explanation:

It tells us it is a combination of substances. We do not know the states of these substances or the state of the new substance. However, we do know that it was physically combined. Hence, this is a mixture. (FYI If it was combined chemically, it would be a compound)

Answer:

Avogadro’s Number

Avogadro’s NumberIt certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small – far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about 10 22 molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only 1.6 × 10 10 molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.

Avogadro’s NumberIt certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small – far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about 10 22 molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only 1.6 × 10 10 molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.Chemists needed a name that can stand for a very large number of items. Amedeo Avogadro (1776 – 1856), an Italian scientist, provided just such a number. He is responsible for the counting unit of measure called the mole. A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for amount of a substance. Just like the dozen and the gross, it is a name that stands for a number. There are therefore 6.02 × 10 23 water molecules in a mole of water molecules. There also would be 6.02 × 10 23 bananas in a mole of bananas, if such a huge number of bananas ever existed

Answer:

Mass = 2000 g

Explanation:

Given data:

Number of molecules of hydrogen = 6.023×10²⁶

Mass of hydrogen = ?

Solution:

1 mole of hydrogen contain 6.022×10²³ molecules

6.023×10²⁶ molecules× 1 mol/ 6.022×10²³ molecules

1.00 ×10³ mol

1000 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 1000 mol × 2 g/mol

Mass = 2000 g

Answer:

[tex]r_{NOCl}=9.28x10^{-2}M[/tex]

Explanation:

Hello!

In this case, given the balanced chemical reaction:

[tex]2 NO(g) + Cl_2(g) \rightarrow 2 NOCl(g)[/tex]

Since there is 1:2 mole ratio between chlorine and NOCl, based on the rate proportions, we can write:

[tex]\frac{1}{-1}r_{Cl_2} =\frac{1}{2}r_{NOCl}[/tex]

It means that for the formation of NOCl, we obtain:

[tex]r_{NOCl}=\frac{2}{-1}r_{Cl_2} \\\\r_{NOCl}=\frac{2}{-1}(-4.64x10^{-2}M)\\\\r_{NOCl}=9.28x10^{-2}M[/tex]

Notice that chlorine is disappearing, which means its rate is negate.

Best regards!

Explanation:

To form bonds with noble gases, a lot of energy is required to form those bonds. Halogens, on the other hand, are extremely reactive. ... The halogens tend to be very reactive, while the noble gases are in no way reactive and don't bond easily, if at all.

Answer:

Explanation:

{H+}*[OH-] = 1 * 10^-14

[2*10^-5]* [OH^-] = 1*10^-14

[OH^-] = 1*10^-14/2*10^-5

[OH^-]  = 5*10^-10

Answer:

 [Ne] 3s² 3d³

Explanation:

According to aufbau's principle of filling electronic orbitals, the sublevels with lower energies are filled up before those with higher energies.

One important to know about this principle is that sublevels do not fill in numerical order.

The order of filling is;

 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f e.tc

From the given choices the wrong one is;

   [Ne] 3s² 3d³

3p fills up before 3d;

Answer: it blue

Explanation:

Answer:

E

Explanation:

From Lu2(CO3)3, we can determine the charge of the variable charge transition metal Lu has a charge of 3, as CO3 has a 2- charge (3*-2 = 6, 6/2 = 3). Then, we apply this to nitrate, which has a charge of 1-. Balance charges, reduce, and you get E.

Answer:

C. The wavelength decreases

Explanation:

This is because frequency is how often the wave hits the top (peak) and the bottom (through). The more it hits the less wavelength it will have because it is moving faster and has more energy.

Copper is higher than silver in the electrochemical series hence copper displaces silver in the electrochemical series.

The solubility of a metal in another is determined by their relative positions in the electrochemical series. The metals that are higher above in the electrochemical series displaces the metals that are lower in the electrochemical series.

Copper is higher than silver in the electrochemical series hence copper displaces silver in the electrochemical series. Therefore, the correct explanation for an observation is that;

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Answer:

Observation 1 is a result of copper ions moving into the solution.

Explanation:

got it right on the test

Answer:

[tex]n_{H_2O}=3.0molH_2O\\\\n_{N_2}=1.0molN_2[/tex]

Explanation:

Hello!

In this case, for the described reaction we have:

[tex]2NH_3(g)+\frac{3}{2} O_2(g)\rightarrow N_2(g)+3H_2O(g)[/tex]

Which means there is 2:3/2 mole ratio between ammonia and oxygen and we use it to compute the consumed moles of ammonia by 13.0 moles of oxygen as shown below:

[tex]n_{NH_3}^{consumed \ by\ O_2}=13.0molO_2*\frac{2molNH_3}{\frac{3}{2}molO_2 } =17.33molNH_3[/tex]

However, since just 2.0 mol of ammonia is available, we infer it is the limiting reactant and the maximum amount of both nitrogen and water that can be produced is computed below:

[tex]n_{H_2O}=2.0molNH_3*\frac{3molH_2O}{2molNH_3} =3.0molH_2O\\\\n_{N_2}=2.0molNH_3*\frac{1molN_2}{2molNH_3} =1.0molN_2[/tex]

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Explanation:

monomer

smaller molecules that are used to prepare a polymer.

may or may not be equivalent to the repeat unit.

oligomer

a molecule consisting of several repeat units of a monomer, but not large enough to be considered a polymer

polymers

arge, usually chainlike molecules that are built from small molecules called monomers. Polymers form the basis for synthetic fibers, rubbers, and plastics and have played a leading role in the revolution that has been brought about in daily life by chemistry.

first synthetic polymers were produced as

by-products of various organic reactions and were regarded as unwanted contaminants.

first completely synthetic plastic

Bakelite, a substance that when molded to a certain shape under high pressure and temperature cannot be softened again or dissolved. Bakelite is a thermoset polymer. In contrast, cellulose nitrate is a thermoplastic polymer; that is, it can be remelted after it has been molded.

ethylene

basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules.

monomer ethylene (C2H4) is

a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of -CH2- units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films).

nylon

the silky appearance and strength of this thread and realized that nylon could be drawn into useful fibers.

The reason for this behavior of nylon is now understood. When nylon is first formed, the individual polymer chains are oriented randomly, like cooked spaghetti, and the sub- stance is highly amorphous. However, when drawn out into a thread, the chains tend to line up (the nylon becomes more crystalline), which leads to increased hydrogen bonding between adjacent chains. This increase in crystallinity, along with the resulting increase in hydrogen-bonding interactions, leads to strong fibers and thus to a highly useful mate- rial. Commercially, nylon is produced by forcing the raw material through a spinneret, a plate containing small holes, which forces the polymer chains to line up.

polyethylene

simplest and one of the best-known polymers, constructed from ethylene monomers.

CH2=CH2

Polyethylene is a member of one subset of synthetic polymers classified as plastics.

properties of polyethylene

Polyethylene is a tough, flexible plastic used for piping, bottles, electrical insulation, packaging films, garbage bags, and many other purposes.

Its properties can be varied by using substituted ethylene monomers. For example, when tetrafluoroethylene is the monomer, the polymer Teflon.

Answer:

a. 167 mL b. 39.27 %

Explanation:

a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol  Al₂(SO₄)₃.

Now, we calculate the number of moles of Al in 45.0 g Al.

We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.

So n = 45.0 g/26.98 g/mol = 1.668 mol

Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol

So, we have 2.5 mol H₂SO₄.

Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.

V = n/C

= 2.5 mol/15.0 mol/L

= 0.167 L

= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.

b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃

Therefore 1.668 mol Al will produce x mol  Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol

So, we need to find the mass of 0.834 mol  Al₂(SO₄)₃. Now molar mass  Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.

Also number of moles of  Al₂(SO₄)₃, n = mass of  Al₂(SO₄)₃,m/molar mass  Al₂(SO₄)₃, M

n =m/M

So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g

% yield = Actual yield/theoretical yield × 100 %

Actual yield = 112 g, /theoretical yield = 285.2 g

So, % yield = 112 g/285.2 g × 100 %

= 0.3927 × 100 %

=  39.27 %

The volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum is 166mL and % yield of the reaction is 39.46%.

Moles of any substance will be calculated by using the below formula as:
n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is :

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

Moles of 45g of Al will be calculated as:

n = 45g / 27g/mol = 1.66 mole

From the stoichiometry of the reaction, it is clear that:

1.66 moles of Al = react with 3/2×1.66=2.49 moles of H₂SO₄

By using the formula of molarity we can calculate the volume of H₂SO₄ as:

M = n/V

V = (2.49) / (15) = 0.166L = 166mL

Again from the stoichiometry it is clear that:
1.66 moles of Al = produces 1/2×1.66= 0.83 moles of Al₂(SO₄)₃

Mass of 0.83 moles of Al₂(SO₄)₃ = (0.83mol)(341.96g/mol) = 283.82 g

Given actual yield of Al₂(SO₄)₃ = 112g

% yield will be calculated as:

Percent yield = (Actual yield/Theoretical yield) × 100

% yield = (112/283.82) × 100 = 39.46%

Hence required values are discussed above.

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Answer:

helium is the smallest and francium is the largest

Answer:

electron pair geometry - octahedral

molecular geometry - square pyramidal

bond angle - < 90 degrees

Explanation:

According to Valence Shell Electron Pair Repulsion Theory (VSEPR), The shapes of molecules depend on the number of electron pairs on the outermost shell of the central atom in the molecule. Recall that electron pairs are always positioned as far apart in space as possible to minimize repulsion.

For a molecule in sp3d2 hybridization, the expected electron domain geometry is octahedral. However, the presence of a lone pair in the molecule distorts the electron pair geometry away from the expected octahedral shape giving a molecular geometry of  square pyramidal and decreases the bond angle less than the expected 90 degrees.

Answer:

Just ask google for help

Explanation:

I always do it and get it right