Question 1
Ernest Rutherford completed his famous gold foil experiment in 1911. In this experiment, alpha particles were fired at a thin sheet of gold foil. He observed that most of the alpha particles passed straight through the gold foil unimpeded, but a small number of alpha particles were deflected. Which of the following conclusions about atomic structure were made from Rutherford’s gold foil experiment?

The amount of \(HClO_3\) that will be produced from 10 liters of \(ClO_2\) would be 8.33 liters.

  \(HClO_3\) is produced from the reaction of  \(ClO_2\) and \(H_2O\) according to the following equation:

\(6 ClO_2 + 3 H_2O --- > 5 HClO_3 + HCl\)

From the balanced equation of the reaction, 6 moles of \(ClO_2\) reacts to produce 5 moles of \(HClO_3\). Thus the mole ratio is 6:5.

Now, assuming the mole ratio also transcribes to the volume ratio and 10 liters of \(ClO_2\) is available for reaction. The amount of \(HClO_3\) that will be produced can be calculated from the volume ratio as follows:

6 moles  \(ClO_2\) = 5 moles  \(HClO_3\)

10 liters  \(ClO_2\) = 5 x 10/6

                       = 8.33 liters

In other words, with 10 liters of \(ClO_2\), the volume of \(HClO_3\) that will be produced would be 8.33 liters.

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Answer:

It can be A or B

Explanation:

Atoms of the same element have the same atomic number so the number of protons (and electrons) are same.

Answer:

1. Molecular equation

BaCl2(aq) + 2AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)

2. Complete Ionic equation

Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯(aq)

3. Net ionic equation

Cl¯(aq) + Ag⁺(aq) —> AgCl(s)

Explanation:

1. Molecular equation:

BaCl2(aq) + AgNO3(aq) –> AgCl(s) + Ba(NO3)2 (aq)

The above equation can be balance as follow:

There are 2 atoms of Cl on the left side and 1 atom on the right side. It can be balance by putting 2 in front of AgCl as shown below:

BaCl2(aq) + AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)

There are 2 atoms of Ag on the right side and 1 atom on the left side. It can be balance by putting 2 in front of AgNO3 as shown below:

BaCl2(aq) + 2AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)

Now, the equation is balanced

2. Complete ionic equation

In solution, BaCl2 and AgNO3 will dissociate as follow:

BaCl2(aq) —> Ba²⁺(aq) + 2Cl¯(aq)

AgNO3(aq) —> Ag⁺(aq) + NO3¯ (aq)

BaCl2(aq) + AgNO3(aq) –>

Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯

3. Net ionic equation

Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯(aq)

Cancel out the spectator ions i.e ions that appears on both side of the equation in order to get the net ionic equation. This is illustrated below:

2Cl¯(aq) + 2Ag⁺(aq) —> 2AgCl(s)

Cl¯(aq) + Ag⁺(aq) —> AgCl(s)

1. Molecular equation: \(BaCl_2(aq) + 2 AgNO_3(aq) \rightarrow 2 AgCl(s) + Ba(NO_3)_2(aq)\)

2. Complete Ionic equation:\(Ba^{2+} (aq)+ 2Cl^- (aq) + 2 Ag^+ (aq) + 2NO_3^-(aq) \rightarrow2AgCl(s) + Ba^{2+}(aq) + 2 NO_3^-(aq)\)3. Net ionic equation: \(Cl^- (aq) + Ag^+ (aq) \rightarrow AgCl(s)\)

1. Molecular equation:

\(BaCl_2(aq) + AgNO_3(aq) \rightarrow AgCl(s) + Ba(NO_3)_2(aq)\)

There are two Cl atoms on the left side and one on the right. Balance can be achieved by adding 2 in front of AgCl, as seen below:

\(BaCl_2(aq) + 2AgNO_3(aq) \rightarrow AgCl(s) + Ba(NO_3)_2(aq)\)

There are two Ag atoms on the right side and one on the left. It is possible to achieve equilibrium by adding 2 in front of \(AgNO_3\), as demonstrated below:

\(BaCl_2(aq) + 2 AgNO_3(aq) \rightarrow 2 AgCl(s) + Ba(NO_3)_2(aq)\)

2. Complete ionic equation

\(BaCl_2\) and \(AgNO_3\) will dissociate in solution as follows:

\(BaCl_2(aq) \rightarrow Ba^{2+}(aq) + 2Cl^-(aq)\)

\(AgNO_3(aq) \rightarrow Ag^+ (aq) + NO_3^-(aq)\)

\(Ba^{2+} (aq)+ 2Cl^- (aq) + 2 Ag^+ (aq) + 2NO_3^-(aq) \rightarrow2AgCl(s) + Ba^{2+}(aq) + 2 NO_3^-(aq)\)

3. Net ionic equation

To obtain the net ionic equation, cancel out the spectator ions, which are ions that appear on both sides of the equation. As an example, consider the following:

\(2Cl^-(aq) + 2Ag^+(aq) \rightarrow 2AgCl(s)\)

\(Cl^-(aq) + Ag^+(aq) \rightarrow AgCl(s)\)

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Answer: The following side effects are common (occurring in greater than 30%) for patients taking Cisplatin:

Nausea and vomiting. Nausea may last up to 1 week after therapy.

Low blood counts.

Kidney toxicity.  

Ototoxicity hearing loss, ringing in the ears.

Blood test abnormalities (low magnesium, low calcium, low potassium)

Explanation:

My mind is ok :')

Answer:

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Explanation:

The equation is given as;

HF-(aq) + H2O(l) --> F-(aq) + H3O(aq)

A bronsted lowry base is any specie that can accept hydrogen ion (proton) from another molecule.

Basically a bronsted lowry base is  a proton acceptor while a bronsted lowry acid is a proton donor.

In the reaction above, upon comparing both the reactants and products;

Bronsted lowry base = Proton acceptor = H2O

Bronsted lowry acid = Proton donor = HF-

Answer:

Hydrogen, lithium, sodium, and potassium all have 1 valence electron ( an electron in the outermost energy level is a valence electron).

a. The chemical equation for the electrolysis of water is:

2H2O(l) → 2H2(g) + O2(g)

b. The gas obtained at the anode during the electrolysis of water is oxygen (O2).

c. The volume of gas obtained at the anode is 0.002232 moles or approximately 0.05 L of oxygen gas.

a. The chemical equation for the electrolysis of water is:

2H2O(l) → 2H2(g) + O2(g)

b. The gas obtained at the anode during the electrolysis of water is oxygen (O2).

c. According to the balanced chemical equation, for every 2 moles of water (H2O) electrolyzed, 1 mole of oxygen gas (O2) is obtained. Since 1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP), we can use the stoichiometry of the reaction to determine the volume of oxygen gas produced.

Given that 50 cm³ of gas is obtained at the anode, we need to convert this volume to liters:

50 cm³ = 50/1000 L = 0.05 L

Using the stoichiometric ratio of the balanced equation, we find that 2 moles of water produce 1 mole of oxygen gas. Therefore, 0.05 L of oxygen gas is equivalent to:

0.05 L × (1 mole/22.4 L) = 0.002232 moles

Thus, the volume of gas obtained at the anode is 0.002232 moles or approximately 0.05 L of oxygen gas.

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Answer:

I guess C.The external parts of the volcano are the summit, slope and magma chamber.

Hope this helps

Answer:

Option C.The external parts of the volcano are the summit, slope and magma chamber.

Explanation:

Because they didn't mention some of the other external parts.  

D. Matter and energy are the same.

Answer:

The answer is

Explanation:

The number of molecules present can be found by using the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 4.21 × 6.02 × 10²³

We have the final answer as

Hope this helps you

Answer:Abundant evidence supports the major contentions of the seafloor-spreading theory. First, samples of the deep ocean floor show that basaltic oceanic crust and overlying sediment become progressively younger as the mid-ocean ridge is approached, and the sediment cover is thinner near the ridge.

Explanation:

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.

Best regards!

Answer:

P4O10

Explanation:

The relation of groundwater and sub-watershed is watershed are the places where the rain water accumulates and this water goes underground and become groundwater.

Groundwater is the most important source of water we have. Groundwater is a part of the watershed.

Thus, the relation of groundwater and sub-watershed is watershed are the places where the rain water accumulates and this water goes underground and become groundwater.

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The calculated formula masses to 133.5 amu, we find that the substance with a formula mass closest to 133.5 amu is (d) AlCl3. Therefore, the answer is option D.

To identify the substance with a formula mass of 133.5 amu, we need to calculate the formula mass of each given substance and compare it to 133.5 amu.

(a) MgCl2:

The formula mass of MgCl2 can be calculated by adding the atomic masses of magnesium (Mg) and chlorine (Cl).

Mg: atomic mass = 24.31 amu

Cl: atomic mass = 35.45 amu

Formula mass of MgCl2 = (24.31 amu) + 2(35.45 amu) = 95.21 amu

(b) SCl:

The formula mass of SCl can be calculated by adding the atomic masses of sulfur (S) and chlorine (Cl).

S: atomic mass = 32.07 amu

Cl: atomic mass = 35.45 amu

Formula mass of SCl = 32.07 amu + 35.45 amu = 67.52 amu

(c) BCl:

The formula mass of BCl can be calculated by adding the atomic mass of boron (B) and chlorine (Cl).

B: atomic mass = 10.81 amu

Cl: atomic mass = 35.45 amu

Formula mass of BCl = 10.81 amu + 35.45 amu = 46.26 amu

(d) AlCl3:

The formula mass of AlCl3 can be calculated by adding the atomic mass of aluminum (Al) and 3 times the atomic mass of chlorine (Cl).

Al: atomic mass = 26.98 amu

Cl: atomic mass = 35.45 amu

Formula mass of AlCl3 = 26.98 amu + 3(35.45 amu) = 133.78 amu. Option D

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Na₂SO₄ contains 1.90834 X 10²⁴ molecules or formula units in a 450 gram sample.

Formula unit- The chemical formula of an ionic compound that lists the ions in the smallest ratio that corresponds to a neutral electrical charge is known as a formula unit. Chemical formulae are used to describe the constituent parts of a compound in chemistry.

Na₂SO₄ molecular mass is 23x2+ 32+ 16X4.

= 142g/mol

Na₂SO₄ mass = 450g

Na₂SO₄ moles are equal to mass/molar mass.

450g/142g/mol = 3.170 mol.

6.02 X 10²³ mol is the Avogadro's number.

formula unit= moles * Avogadro's number

⇒3.17 X 6.02 X 10²³ molecules= 1.90834 X 10²⁴  

Na₂SO₄ therefore contains 1.90834 X 10²⁴ molecules or formula units..

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4.5 moles of O2 would make 3.40 moles of CO2.

The steps for this process are as follows

Step 1: Balance the equation:

C3H8 + 5O2 → 3CO2 + 4H2O

Step 2: Determine the molar ratio between O2 and CO2.

The molar ratio between O2 and CO2 is 5:3.

Step 3: Calculate the moles of CO2 produced.

If you start with 4.5 moles of O2, you can calculate the moles of CO2 produced using the molar ratio:

4.5 moles O2 × (3 moles CO2/5 moles O2) = 2.7 moles CO2

Since you can't have a fraction of a mole, we have to round the answer up to the nearest whole number.

This means that 4.5 moles of O2 will produce 3.75 moles of CO2.

In the given question,  in the options not give 3.75 moles, but have 3.40 mole , so the answer is 3.40 moles its the approximate value of 3.75

This question uses the concept of stoichiometry, which is the calculation of the amount of substances involved in a chemical reaction.

By knowing the molecular ratios between the reactants and products, it is possible to calculate the amount of each product that is produced from a given amount of reactants.

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Silicon has intermediate properties between carbon and germanium, since its atomic number is 14.

Because it is part of the metalloids, silicon is resistant to most acids and reacts with dilute alkalis and halogens. These properties allow it to have many uses, for example:

- Brick production (from silicon dioxide).

- Fertilizer (from silicon silicates).

- Manufacture of glass and insulation, and electrical circuit parts (because it is a semiconductor).

So, silicon is very versatile in addition to being a very abundant material.

This molecule in the question presents 6 Carbons in its main chain, counting only the carbons in the middle, which makes it a Hexane molecule. Besides the main chain carbons, we also have two methyl groups linked to carbons 2 and 3, counting from left to right. Therefore the final name will be:

2,3-dimethylhexane. Letter C

Answer:

69.7 g of Na₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

Next, we shall determine the masses of H₂SO₄ and NaOH that reacted and the mass of Na₂SO₄ produced from the balanced equation. This can be obtained as shown below:

Molar mass of H₂SO₄ = (2×1) + 32 + (4×16)

= 2 + 32 + 64

= 98 g/mol

Mass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mass of NaOH from the balanced equation = 2 × 40 = 80 g

Molar mass of Na₂SO₄ = (2×23) + 32 + (4×16)

= 46 + 32 + 64

= 142 g/mol

Mass of Na₂SO₄ from the balanced equation = 1 × 142 = 142 g

SUMMARY:

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH to produce 142 g of Na₂SO₄.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH.

Therefore, 48.1 g of H₂SO₄ will react with = (48.1 × 80)/98 = 39.3 g of NaOH.

From the calculation made above, we can see that only 39.3 g of NaOH out of 51 g given was needed to react completely with 48.1 g of H₂SO₄. Therefore, H₂SO₄ is the limiting reactant and NaOH is the excess reactant.

Finally, we shall determine the maximum mass of Na₂SO₄ produced from the reaction. This can be obtained by using the limiting reactant as illustrated below:

NOTE: H₂SO₄ is the limiting

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄.

Therefore, 48.1 g of H₂SO₄ will react to produce = (48.1 × 142)/98 = 69.7 g of Na₂SO₄.

Thus, 69.7 g of Na₂SO₄ were obtained from the reaction.

Answer:

69.23g

Explanation:

Find out how many moles is in 9.89g of H2.

number of moles = mass(g) / molar mass

1 is the molar mass of hydrogen (to the nearest whole)

relative molecular mass of H2: 2*1 = 2

number of moles of H2 = 9.89/2 = 4.945

1 mol of H2 is produced from 2 mol of Li

so

4.945 mol of H2 produces 9.89 mol of Li

mass(g) = number of moles * molar mass

7 is the molar mass of Lithium (to the nearest whole)

mass = 9.89 * 7 = 69.23

69.23 grams of Li are needed to produce 9.89 of H2

Answer:

27 days, 7 hours and 43 minutes.

Explanation:

To complete a full orbit around Earth, it takes 27 days, 7 hours and 43 minutes for the Moon. This is called sidereal month and is measured by the Moon's position relative to distant *fixed" stars.

The oxidation number for carbon in H2CO3 is +4, is so option d.

It is the electrical charge of an atom when it is part of a compound. This charge depends on the number of electrons gained, lost or shared by an atom.

                               H   C  O

Oxidation number: +1 +4 -2

Carbon (Z=6) can have oxidation numbers ranging from -4 to +4.

C can present different oxidation states within the same compound that contains more than one atom of it.

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Answer:

B) It will increase.

Explanation:

Solubility tends to increase with an increase in temperature

Answer: B

Explanation: Anything that has mass and takes up space

The research setup is a comparative analysis of several boats in an effort to understand their performance on the water and the differences between them because little would be learned from studying just one boat. Seven boats have been chosen for additional examination. They are all either "traditional" boats or boats that have been found in archaeology. Smaller open boats are the main focus.

The mass of an object per unit volume is what is meant by density. A brick, for instance, weighs more than a Styrofoam block of equal size (volume), indicating that the brick is denser than the Styrofoam. An object will float in a liquid if it has a lower density than the liquid. An object will sink in a liquid if it has a higher density than the liquid.

Instruct students to work in teams of two. First, they design their boats, keeping in mind that the boats must float while holding weight and that they have a limited amount of clay to use. Students can test their designs and make changes as often as they like. The goal is to create boats that can hold as much weight as possible while staying afloat.

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Answer:

3KOH(aq) + FeCl3(aq) --------> Fe(OH)3 + 3KCl

Explanation:

The rule for balancing chemical equations is that the number of atoms of each element must be the same on both sides of the reaction equation.

Hence, the two reactants are potassium hydroxide and aqueous iron(III) chloride.

The balanced molecular reaction equation is;

3KOH(aq) + FeCl3(aq) --------> Fe(OH)3 + 3KCl

The steps required to determine the empirical formula of a compound from the mass % are:

1. Convert the mass percentages of each element to moles using the molar mass formula.

2. Divide the mass percentages among the smallest number to calculate the ratio of atoms of each element present.

3. Ensure that the ratio of atoms of each element is a ratio of integer numbers.

The empirical formula of a compound is the simplest whole number ratio of the elements present in the compound. It is often represented using the symbols of the elements involved and can be calculated from the relative masses of the elements. It is helpful in identifying the composition of a compound. For example, the empirical formula for glucose is CH2O, indicating that there is one carbon atom, two hydrogen atoms, and one oxygen atom present in the compound.

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The mass of 1.50 moles of sodium chloride is 87.66 grams, which is calculated by multiplying the number of moles (1.50) by the molar mass of sodium chloride (58.44 g/mol).

To calculate the mass in grams of 1.50 moles of sodium chloride, you first need to determine the molar mass of sodium chloride (NaCl). Since sodium (Na) has a molar mass of 22.99 g/mol, and chlorine (Cl) has a molar mass of 35.45 g/mol, you can add these together to find that the molar mass of sodium chloride is 58.44 g/mol. Then, you simply multiply the number of moles (1.50) by the molar mass (58.44 g/mol) to find that 1.50 moles of sodium chloride have a mass of 87.66 grams.

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