Ernest Rutherford's experiments in which alpha rays pass through a thin piece of metal foil demonstrated that atom contains a tiny nucleus containing most of the atom's mass. The correct answer choice is "the atom contains a tiny nucleus containing most of the atom's mass."
The alpha-particle scattering experiment was performed by Ernest Rutherford in 1911. It was conducted to discover the nature of atomic structure. The experiment demonstrated that most of the mass of an atom and all of its positive charge are contained in a small nucleus at the center of the atom.
The electrons were found to occupy almost all of the remaining space. Therefore, the correct answer is the atom contains a tiny nucleus containing most of the atom's mass.
Therefore "the atom contains a tiny nucleus containing most of the atom's mass." is the correct answer.
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which of the following is a characteristic of fractional distillation? select all that apply: components with higher boiling points tend to rise to the top of the distillation apparatus. repeated vaporization and condensation are part of the fractional distillation process. fractional distillation is useful for the separation of liquids with boiling points that are close together. the apparatus used for fractional distillation is often a multi-level insulated column.
The components with higher boiling point will be at the bottom during fractional distillation. So, all of the following statement follows except option (A).
Fractional distillation process is defined as the process of separation of a mixture into its component parts or into the fractions. Chemical compounds are separated by heating them to a temperature after which one or more fractions of the mixture will vaporize from their component. This process uses distillation to fractionate. It is a process by which components in a chemical mixture are separated into different parts according to their different boiling points. This distillation process is used to purify chemicals and to separate mixtures to obtain their components. This involves repeated vaporization and condensation are part of the fractional distillation process.
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The complete question is,
which of the following is a characteristic of fractional distillation? select all that apply:
A. components with higher boiling points tend to rise to the top of the distillation apparatus.
B. repeated vaporization and condensation are part of the fractional distillation process.
C. fractional distillation is useful for the separation of liquids with boiling points that are close together.
D. the apparatus used for fractional distillation is often a multi-level insulated column.
2.98x10^7+3.12x10^7 and a expressed answer in scientific notation
When adding numbers in scientific notation, we need to ensure that the exponents of 10 are the same.
2.98x10^7 + 3.12x10^7 can be rewritten as:
(2.98 + 3.12) x 10^7
= 6.10 x 10^7
Therefore, the sum of 2.98x10^7 and 3.12x10^7 in scientific notation is 6.10x10^7.
a buffer is made by combining 20.0 ml 0.250 m nh4cl with 30.0 ml 0.250 m nh3. a. calculate the ph of the buffer. 14 b. calculate the ph of the buffer after 1.00 ml of 6.00 m hcl is added and equilibrium is re- established. c. calculate the ph of the buffer after 1.00 ml of 6.00 m naoh is added to a fresh sample of the buffer and equilibrium is re-established.
a. The pH of the buffer is 9.24.
b. The pH of the buffer after the addition of HCl is 8.68.
c. The pH of the buffer after the addition of NaOH is 9.37.
The chemical equation for the reaction between NH₄⁺ and NH₃ is,
NH₄⁺ + NH₃ ⇌ NH₃ + NH₄⁺.
At equilibrium, the concentration of NH₄⁺ equals the concentration of NH₃. The Ka for NH₄⁺ is 5.6 × 10^-10.
Using the Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), where A⁻ is NH₃ and HA is NH₄⁺, we can calculate the pH of the buffer as follows:
pH = pKa + log([A⁻]/[HA])
pH = pKa + log([NH₃]/[NH₄⁺])
pH = pKa + log([0.250 M]/[0.250 M])
pH = -log(5.6 × 10^-10) + log(1)
pH = 9.24
When 1.00 mL of 6.00 M HCl is added to the buffer solution, it reacts with NH3 to form NH4+ and Cl^- ions.
The new concentration of NH₄⁺ is
[NH₄⁺] = 0.250 M + (1.00 × 10^-3 L)(6.00 M)/(20.0 mL + 1.00 mL) = 0.302 M.
The new concentration of NH₃ is [NH₃] = 0.250 M - (1.00 × 10^-3 L)(6.00 M)/(30.0 mL + 1.00 mL) = 0.196 M. Using the Henderson-Hasselbalch equation as before, we can calculate the new pH of the buffer as follows:
pH = pKa + log([A⁻]/[HA])
pH = pKa + log([NH₃]/[NH₄⁺])
pH = -log(5.6 × 10^-10) + log(0.196/0.302)
pH = 8.68
When 1.00 mL of 6.00 M NaOH is added to a fresh sample of the buffer, it reacts with NH₄⁺ to form NH₃ and Na+.
The new concentration of NH₄⁺ is,
[NH₄⁺] = 0.250 M - (1.00 × 10^-3 L)(6.00 M)/(20.0 mL + 1.00 mL)
= 0.198 M.
The new concentration of NH₃ is [NH₃] = 0.250 M + (1.00 × 10^-3 L)(6.00 M)/(30.0 mL + 1.00 mL) = 0.304 M. Using the Henderson-Hasselbalch equation as before, we can calculate the new pH of the buffer as follows:
pH = pKa + log([A⁻]/[HA])
pH = pKa + log([NH₃]/[NH₄⁺])
pH = -log(5.6 × 10^-10) + log(0.304/0.198)
pH = 9.37
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You are given the mass of one of the reactants in a chemical reaction and asked to find the mass of one of the products. What is the first step? Question 3 options: check to make sure the equation is balanced make the switch using the mole ratio give up get out the balance
Answer:
Check to make sure the equation is balanced
Explanation:
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what is the probability of finding a hydrogen 1s electron within the so called van der waals radius of hydrogen
The probability of finding a hydrogen 1s electron within the so-called van der Waals radius of hydrogen is 90%.
Van der Waals radius (vdW) is an estimate of the size of an atom. It is the radius of a sphere encompassing the outermost orbital of an atom (such as van der Waals radii).The probability of locating an electron in a particular area is referred to as probability density. The probability density of an electron in the hydrogen atom in its ground state is greatest at the center of the atom and decreases exponentially as the distance from the nucleus increases.The probability of finding a hydrogen 1s electron within the so-called van der Waals radius of hydrogen is approximately 90%.
The radius of the hydrogen atom is 53 pm, and the van der Waals radius of the hydrogen atom is 120 pm, respectively. As a result, there is a higher likelihood of locating a hydrogen 1s electron within the van der Waals radius of hydrogen than outside of it, which is roughly 90%.
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45. Perchloric acid (HCIO) reacts with aqueous potassium carbonate, forming carbon dioxide gas and water.
Answer:
2 HClO4 + K2CO3 → CO2 + 2 KClO4 + H2O
Explanation:
The balanced chemical equation for the reaction between perchloric acid (HClO4) and aqueous potassium carbonate (K2CO3) can be written as follows:
2 HClO4 + K2CO3 → CO2 + 2 KClO4 + H2O
In this reaction, two molecules of perchloric acid react with one molecule of aqueous potassium carbonate to produce one molecule of carbon dioxide gas, two molecules of potassium perchlorate, and one molecule of water.
Note that this reaction is a double displacement reaction, also known as a metathesis reaction, where the cations and anions of two different compounds exchange places, forming two new compounds. In this case, the hydrogen cation (H+) and the potassium cation (K+) exchange places, while the perchlorate anion (ClO4-) and the carbonate anion (CO3^2-) exchange places.
the haber process synthesizes ammonia at elevated temperatures and pressures from nitrogen gas and hydrogen gas. what volume of ammonia would be produced if 350 l of nitrogen gas and 900 l of hydrogen gas is combined at stp? assume the reaction goes to completion.
When 350 L of nitrogen gas and 900 L of hydrogen gas are combined at STP using the Haber process, 329 L of ammonia gas is produced.
Assuming the reaction goes to completion, the Haber process synthesizes ammonia at elevated temperatures and pressures from nitrogen gas and hydrogen gas. As a result, balanced chemical equation can be given as:N2(g) + 3H2(g) → 2NH3(g)The balanced equation establishes that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.
This implies that the stoichiometric ratios of the reactants are 1:3. Thus, if 350 L of nitrogen and 900 L of hydrogen are mixed in a reaction vessel and the reaction proceeds to completion, the limiting reactant will be nitrogen. Hence, the number of moles of nitrogen present is calculated using the ideal gas equation n = PV/RT as follows:n(N2) = PV/RT = (1 atm) x (350 L) / (0.08206 L atm/mol K) x (273 K) = 14.07 mol
Similarly, the number of moles of hydrogen is calculated as:n(H2) = PV/RT = (1 atm) x (900 L) / (0.08206 L atm/mol K) x (273 K) = 36.25 molSince nitrogen is the limiting reactant, it will completely react with 1/3 of the amount of hydrogen present to produce ammonia. As a result, the amount of ammonia generated will be equivalent to the quantity of nitrogen that reacted. Therefore, the volume of ammonia is calculated as follows:n(NH3) = n(N2) = 14.07 molV(NH3) = n(NH3) x (RT/P) = 14.07 x (0.08206 x 273 / 1) / (1) = 329 L
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pickles are made by immersing cucumbers in a concentrated saltwater solution. explain what happens to the cucumber in this process to cause it to shrink and taste salty.
When cucumbers are immersed in a concentrated saltwater solution, the process of osmosis occurs. Osmosis is the movement of water molecules across a selectively permeable membrane from an area of high water concentration to an area of low water concentration.
In this case, the concentrated saltwater solution outside the cucumber has a much lower water concentration than inside the cucumber. As a result, water from inside the cucumber moves out of the cell membrane and into the saltwater solution, causing the cucumber to shrink.
Additionally, the salt ions from the solution enter the cucumber through the cell membrane, which is permeable to ions. The presence of salt ions in the cucumber affects the taste, making it salty. The salt also acts as a preservative, inhibiting the growth of microorganisms that can spoil the cucumber. This is why pickles have a longer shelf life than fresh cucumbers.
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To prepare 100.0 mL of 0.0300 M HCl, what volume of 0.500 M stock solution is required?
6.00 mL of the 0.500 M HCl stock solution is required to prepare 100.0 mL of 0.0300 M HCl.
We can use the following formula to calculate the volume of stock solution needed:
V(stock) x M(stock) = V(final) x M(final)
where V(stock) is the volume of the stock solution, M(stock) is the concentration of the stock solution, V(final) is the final volume of the diluted solution, and M(final) is the desired concentration of the diluted solution.
Plugging in the given values, we get:
V(stock) x 0.500 M = 100.0 mL x 0.0300 M
Solving for V(stock), we get:
V(stock) = (100.0 mL x 0.0300 M) / 0.500 M = 6.00 mL
A diluted solution is a solution that has been made weaker by adding more solvent (usually water) to a more concentrated solution. This results in a decrease in the concentration of the solute in the solution.
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is it true that in a chemical reaction new types of atoms that are different from those of the reactants are produced to form new substances
No, it is not true that new types of atoms are produced in chemical reaction.
What is meant by chemical reaction?A chemical reaction is a process that results in the chemical conversion of one group of chemical compounds into another.
Chemical reactions involve the rearrangement of atoms to form new substances, but atoms themselves are not created or destroyed in the process. This is known as law of conservation of mass, which states that the total mass of reactants in a chemical reaction is equal to the total mass of products.
Therefore, the types of atoms present in the reactants are also present in the products, just in different combinations and arrangements.
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at what ph must the solution in the lead-acid battery be to produce its standard cell potential of 2.05 v?
At ph = 0 must the solution in the lead-acid battery be to produce its standard cell potential of 2.05 v.
Pb + PbO₂ + 2H₂SO₄ ⇄ 2PbSO₄ + 2H₂O
At anode: Pb + SO²⁻₄ → PbSO₄ + 2e⁻
E = 0.356 V
At Cathode: PbO₂ + H₂SO₄ +2H⁺ + 2e⁻→ PbSO₄ +2H₂O
E = 1.690 V
Overall:
Pb + PbO₂ + 2H₂SO₄ ⇄ 2PbSO₄ + 2H₂O
E =2.05 at standard concentration = 1 M
ph = log[1]
ph= 0
The cell potential is the voltage of a single electrochemical cell, to put it simply. To boost the voltage of the battery, many cells may be packaged in series. A completely charged lead-acid battery will always have a cell potential of about 2.1 V, regardless of the battery's size since cell potential is a property of a specific chemical reaction. Similar to this, depending on the cathode material, lithium-ion batteries have nominal cell potentials between 3.2 and 3.85 V.
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