Answer:
Explanation:
dilute solutions of hydrochloric acid (HCl), sulphuric acid (H₂SO₄), and nitric acid (HNO₃) react with active metals to produce a salt and hydrogen gas.
Active metals react strongly and quickly with other elements and compounds due to the electrons in its structure and its ease of sharing the electrons with other elements.
The most active metals are found in Groups 1 and 2 of the Periodic Table (i.e. the left side), and include lithium, potassium, magnesium, and calcium. Metals such as aluminium, lead, and zinc, are less active than magnesium or calcium, but are generally still labelled as 'active'. Metals such as copper, gold, or silver are inactive and will not react.
Therefore, in the provided question, all the metals listed, except for copper, will produce a metal salt + hydrochloric acid. Copper will not react.
In general:
metal + HCl = metal chloride + H₂ metal + H₂SO₄ = metal sulphate + H₂metal + HNO₃ = metal nitrate + H₂What is true of spontaneous reactions?
O They are indicated by a negative change in Gibbs free energy.
O They have a positive value of AS.
O They are instantaneous.
O They always release heat.
Help 20pts
In the Periodic Table below, shade all the elements for which the neutral atom has an outer electron configuration of ms2nd2, where n and m are integers, and =m+n1.
The elements that have an outer electron configuration of ms2nd2 are located in the d-block of the periodic table and include some of the transition metals and lanthanides.
What is the periodic table?To determine which elements in the periodic table have this outer electron configuration, you can look at the position of the d-block elements in the table. The d-block elements are located in the middle of the table and include the transition metals. These elements have partially filled d orbitals, which can accommodate up to 10 electrons.
Elements in the d-block with an atomic number of 21 through 30 (scandium through zinc) have an outer electron configuration of d10s2 and do not fit the ms2nd2 configuration. However, elements in the d-block with an atomic number of 39 through 48 (yttrium through cadmium) have an outer electron configuration of d10s2p1 and can have the ms2nd2 configuration by removing the single electron in the p orbital. Elements in the d-block with an atomic number of 57 through 80 (lanthanum through mercury) also have the possibility of having an outer electron configuration of ms2nd2.
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CaCO3 + 2HCI =CaCl2 + H₂O + CO2
5. Calcium carbonate (CaCO3) combines with HCl to produce calcium chloride (CaCl₂),
water, and carbon dioxide gas (CO₂). How many grams of HCI are required to react with
6.35 mol CaCO3?
463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.
What is meant by molar mass?Mass of one mole of substance is referred to as the molar mass. The molar mass of a substance can be calculated by adding up the atomic masses of all the atoms in a molecule.
Balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
6.35 mol CaCO₃ * 2 mol HCl / 1 mol CaCO₃ = 12.7 mol HCl
Now, we use the molar mass of HCl (36.46 g/mol) to convert from moles to grams: 12.7 mol HCl * 36.46 g/mol = 463.5 g HCl
Therefore, 463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.
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6. What is the pH of a 0.25 M solution of NH4Cl? [Kb(NH3) = 1.8 10–5
The Ammonium Chloride solution at 0.25 M has a pH of 2.67.
Why is the pH of Ammonium Chloride below 7?As a result, the weak basic (Chlorine) in the solution is overpowered by the conjugate acid (Ammonium cation), making the solution mildly acidic. According to the equation pH =log[Hydrogen ion], an acidic solution has a pH lower than 7. Aqueous ammonium chloride solution has a pH that is less than 7.
Ammonium cation + Water ⇌ Nitrogen trihydride + Hydronium ion
Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]
[Nitrogen trihydride] = [Hydronium ion] = x
[Ammonium cation] = 0.25 - x
Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]
1.8 × 10–5 = x² / (0.25 - x)
1.8 × 10–5 = x² / 0.25
x² = 4.5 × 10–6
x = 2.12 × 10–3
pH = -log[Hydronium ion] = -log(2.12 × 10–3) = 2.67
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CHALLENGE The circles below represent of the large circle, and multiply it by 30. That Earth and the moon. Measure the diameter would be the correct distance from Earth to the moon at this scale. Draw the two circles in the space provided. Use the correct distance you found.● = Earth ●=moon
To draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.
What is the explanation for the above response?If we assume that the larger circle represents the Earth, then the diameter of the Earth would be 30 times the diameter of the smaller circle representing the moon. Let's say that the diameter of the smaller circle is x. Then the diameter of the larger circle (Earth) would be 30 times x or 30x.
To find the correct distance from Earth to the moon at this scale, we need to know the actual distance from Earth to the moon, which is approximately 238,855 miles or 384,400 kilometers. If we divide this distance by the scale factor of 30, we get:
238,855 miles / 30 = 7,961.8 miles
Therefore, the diameter of the smaller circle (moon) would be approximately 7,961.8 miles / π = 2,532.5 miles (rounded to one decimal place). And the diameter of the larger circle (Earth) would be 30 times that or 75,974.4 miles
So, to draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.
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Draw both enantiomers of the following compound
Enantiomers rotate the plane of polarized light in opposite directions, and this property is used to distinguish between them in a process called optical rotation.
What are the enantiomers of a compound?Enantiomers are pairs of molecules that are non-superimposable mirror images of each other.
They are isomers, meaning they have the same molecular formula and connectivity but differ in their three-dimensional arrangement of atoms in space.
Enantiomers exhibit identical physical and chemical properties, except for their interaction with plane-polarized light (a type of light that oscillates in a single plane).
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2. A student prepared a 0.500 M solution of an unknown acid, and measured the pH as 3.56 at 25°C. (a) What is the acid dissociation constant of this unknown acid? (b) What percentage of acid is ionised in this solution
To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution.
(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:
[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M
What is the acid dissociation constant of this unknown acid?The acid dissociation constant (Ka) can then be calculated using the equation:
Ka = [H+][A-]/[HA]
where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:
α = [A-]/[HA]
Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:
[A-] ≈ 0.500α
[HA] ≈ 0.500 - 0.500α
Substituting these expressions into the equation for Ka, we get:
Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α
≈ ([H+]/Ka)(0.500α)/(1-α)
Solving for Ka, we get:
Ka ≈ H+/0.500α
Substituting the values we have calculated, we get:
Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)
Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).
(b) To find the percentage of acid that is ionized in the solution, we can use the equation:
α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))
where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:
α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008
Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.
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To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution.
(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:
[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M
What is the acid dissociation constant of this unknown acid?The acid dissociation constant (Ka) can then be calculated using the equation:
Ka = [H+][A-]/[HA]
where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:
α = [A-]/[HA]
Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:
[A-] ≈ 0.500α
[HA] ≈ 0.500 - 0.500α
Substituting these expressions into the equation for Ka, we get:
Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α
≈ ([H+]/Ka)(0.500α)/(1-α)
Solving for Ka, we get:
Ka ≈ H+/0.500α
Substituting the values we have calculated, we get:
Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)
Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).
(b) To find the percentage of acid that is ionized in the solution, we can use the equation:
α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))
where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:
α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008
Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.
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A flask filled to the 25.0 ml mark contain 29.97 g of a concentrated salt water solution. What is the density of the solution?
A concentrated saltwater solution weighing 29.97 g and fitting into a flask to the mark of 25.0 ml has a density of about 1199.2 g/L.
How is the density of the solution determined?By dividing the solution's mass by its volume, we may get its density: density = mass/volume
We need to know the density of water at the solution's temperature as well as the capacity of the flask up to the 25.0 ml level in order to calculate the volume of the solution.
Since 1 mL = 0.001 L, volume is equal to 25.0 mL, or 0.0250 L.
Now, we may determine the solution's density as follows:
1199.2 g/L or 29.97 g/0.0250 L is what is referred to as density.
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What is eutectic temperature
The eutectic point is the lowest temperature at which the liquid phase is constant at a particular pressure.
What does the word "eutectic" mean?A melting composition known as a eutectic consists of at least two components that melt and freeze at the same rates. The components combine during the crystallisation phase, operating as a single component as a result.
What are eutectic pressure and temperature?The eutectic is the system's lowest melting point under its own pressure; it has a matching temperature called the eutectic temperature and produces the eutectic liquid as a result. In terms of composition, eutectic liquids are located between the system's solid phases.
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pls help!!!
a compound is found to be 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen. it has a molecular molar mass of 140.22 g/mol. what is the molecular formula.
show work pls!!
The molecular formula of the compound, given that it contains 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen is C₆H₁₂N₄
How do i determine the molecular formula?To obtain the molecular formula, we must first determine the empirical formula. Details on how to obtain the empirical formula is given beloww:
Carbon (C) = 51.39%Hydrogen (H) = 8.64%Nitrogen (N) = 39.97%Empirical formula =?Divide by their molar mass
C = 51.39 / 12 = 4.283
H = 8.64 / 1 = 8.64
N = 39.97 / 14 = 2.855
Divide by the smallest
C = 4.283 / 2.855 = 1.5
H = 8.64 / 2.855 = 3
N = 2.855 / 2.855 = 1
Multiply through by 2 to express in whole number
C = 1.5 × 2 = 3
H = 3 × 2 = 6
N = 1 × 2 = 2
Thus, we can conclude that the empirical formula is C₃H₆N₂
Finally, we shall determine the molecular formula. Details below
Empirical formula = C₃H₆N₂Molar mass of compound = 140.22 g/molMolecular formula =?Molecular formula = empirical × n = mass number
[C₃H₆N₂]n = 140.22
[(12×3) + (1×6) + (14×2)]n = 140.22
70n = 140.22
Divide both sides by 70
n = 140.22 / 70
n = 2
Molecular formula = [C₃H₆N₂]n
Molecular formula = [C₃H₆N₂]₂
Molecular formula = C₆H₁₂N₄
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At 25 ∘C
, the equilibrium partial pressures for the reaction
A(g)+2B(g)↽−−⇀C(g)+D(g)
were found to be A=5.63
atm, B=5.00
atm, C=5.47
atm, and D=5.63
atm.
What is the standard change in Gibbs free energy of this reaction at 25 ∘C
?
The standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.
What is standard change?
To find the standard change in Gibbs free energy of the reaction, we need to use the following equation:
ΔG° = -RT ln(K)
where ΔG° is the standard change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25 °C = 298 K), and K is the equilibrium constant.
To find K, we need to use the equilibrium partial pressures:
K = (PC × PD) / (PA × PB²)
where PA, PB, PC, and PD are the equilibrium partial pressures of A, B, C, and D, respectively.
Substituting the values, we get:
K = (5.47 atm × 5.63 atm) / (5.63 atm × (5.00 atm)²)
K = 0.6176
Now we can calculate the standard change in Gibbs free energy:
ΔG° = -RT ln(K)
ΔG° = -(8.314 J/mol·K) × (298 K) × ln(0.6176)
ΔG° = -1,690 J/mol or -1.69 kJ/mol
Therefore, the standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.
What is free energy?
Free energy, also known as Gibbs free energy, is a thermodynamic quantity that represents the amount of energy in a system that is available to do work at a constant temperature and pressure. It is denoted by the symbol G and is expressed in units of joules (J) or calories (cal).
In simple terms, free energy is the energy that can be used to do work. It is defined by the equation:
ΔG = ΔH - TΔS
where ΔH is the change in enthalpy (heat content) of the system, ΔS is the change in entropy (disorder) of the system, and T is the absolute temperature in Kelvin.
If ΔG is negative, the reaction is spontaneous and can proceed without the input of external energy. If ΔG is positive, the reaction is non-spontaneous and requires energy input to proceed. If ΔG is zero, the system is at equilibrium.
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2AI + 6HCI=2AlCl3 + 3H₂
3. Aluminum reacts with HCI to produce aluminum chloride (AICI3) and hydrogen gas (H₂).
Calculate the number of moles of HCI required to react with 0.62 moles of Al.
3.0 moles of [tex]Al[/tex] can fully react with hydrogen chloride to produce 4.5 moles of [tex]H_{2}[/tex]. Thus, 0.93 moles will be produced by 0.62 moles of [tex]Al[/tex].
STOICHIOMETRYBased on this inquiry, how does aluminum react with hydrogen chloride to produce aluminum chloride and hydrogen gas[tex]Al +6HCl= AlCl_{3} +3H_{2}[/tex]According to this equation, 3 moles of hydrogen gas are produced during the reaction of 2 moles of aluminum ([tex]Al[/tex]).As a result, 3 moles of aluminum will result in 3 3 2 = 4.5 moles of hydrogen gas.As a result, the entire reaction of 3.0 moles of [tex]Al[/tex]with hydrogen chloride can produce 4.5 moles of [tex]H_{2}[/tex].The proportion of reactants to products before, during, and after chemical processes is known as stoichiometry.For more information on stoichiometry kindly visit to
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If you started with 20.0 g of a radioisotope and waited for 3 half-lives to pass, then how much would remain? 2.50 g 5.00 g 10.0 g 15.0 g
The amount that would remain, given that 3 half-lives has pass when you started with 20.0 g is 2.50 grams (1st option)
How do i determine the amount that would remain?The following data were obtained from the question:
Original amount of radioisotope (N₀) = 20.0 gramsNumber of half-lives that has passed (n) = 3Amount remaining after 3 half-lives (N) = ?The amount remaining can be obtained as shown below:
N = N₀ / 2ⁿ
N = 20 / 2³
N = 20 / 8
N = 2.50 grams
Thus, we can conclude from the above calculation that the amount that would remain after 3 half-lives to pass is 2.50 grams (1st option)
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Answer:
2.50g
Explanation:
Round to 2 significant
figures.
5,249
5,250. The number was rounded up from 5,249 because the last digit, 9, is greater than or equal to 5.
What is rounded up?Rounding up is a mathematical operation that involves increasing a number to its nearest whole number. It is commonly used when dealing with money, measurements, or statistics. When rounding up, the number is increased to the next highest whole number. For example, if a number is 6.7, it would be rounded up to 7. Rounding up is often used when dealing with exact measurements or estimates to simplify the calculations. It can also be used to make the results of a calculation easier to understand. In the case of money, rounding up can be used to round a number to the nearest dollar. This prevents dealing with fractional amounts of money. Rounding up can also be utilized in statistical analysis, such as in the calculation of mean or median. This simplifies the data and prevents dealing with fractions or decimals.
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Question 5(Multiple Choice Worth 3 points)
(07.02 LC)
The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of thermal energy. Which one do you expect to increase in temperature the least?
a) Cadmium, 0.230 J/(g °C)
b) Sodium, 1.21 J/(g °C)
c) Water, 4.184 J/(g °C)
d) Hydrogen, 14.267 J/(g °C)
Component form of the vector v is as follows: 4 3 1.5 1 Using the standard basis vectors I and j), express the vector w as follows: 3 two 1 4 pp . 1 3 w 3.5 C. V plus w= d. Determine the vector v's magnitude
What does "vector" mean?
Latin word for "carrier" is "vector." Point A is transported to point B by vectors. The orientation of the vectors AB is the direction in which point A is moved in relation to point B, and the amplitude of the vector is the width of the line connecting the two locations A and B. The terms Euclidean vectors and spatial vectors are also used to refer to vectors.
A vector space is what?
A vector space, also known as a linear space, is a collection of things called vectors that can be added to and multiplied ("scaled") by figures called scalars in the fields of mathematics, physics, and engineering.
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The calcium and magnesium in a urine sample were precipitated as oxalates. A mixed precipitate of calcium oxalate (CaC2O4) and magnesium oxalate (MgC2O4) resulted and was analysed by gravimetry. The formed precipitate mixture was heated to form calcium carbonate (CaCO3) and magnesium oxide (MgO) with a total mass of 0.0433 g. The solid precipitate mixture was ignited to form CaO and MgO, the resulting solid after ignition weighed 0.0285 g. What was the mass of calcium in the original sample? All answers should be reported with the correct significant figures
The mass of calcium in the original urine sample would be 0.0140 g.
Stoichiometric problemFirst, we need to find the masses of calcium and magnesium oxalates in the original sample. Let x be the mass of calcium oxalate and y be the mass of magnesium oxalate. Then we have:
x + y = mass of the mixed oxalate precipitate
Next, we need to use the information given to find the mass of calcium in the original sample. The mass of calcium oxide formed after ignition is equal to the mass of calcium oxalate in the original sample. We can calculate the mass of calcium oxide using the mass of calcium carbonate formed and the molar mass ratio of calcium carbonate to calcium oxide.
The balanced chemical equations for the reactions are:
CaC2O4 -> CaCO3 + CO2
CaCO3 -> CaO + CO2
The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CaO is 56.08 g/mol.
From the given information, we have:
0.0433 g = (x + y)(100.09 g/mol + 80.15 g/mol) / (128.10 g/mol + 80.15 g/mol)
0.0285 g = x(56.08 g/mol) + y(40.31 g/mol)
Solving these equations simultaneously, we get:
x = 0.0140 g
y = 0.0053 g
Therefore, the mass of calcium in the original sample (which is equal to the mass of calcium oxide formed after ignition) is:
0.0140 g
So the mass of calcium in the original sample is 0.0140 g.
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The two possible units of molarity are
Answer: The units for molarity are moles/liter.
Similarly, the equation to find molarity is moles divided by liters.
Explanation:
mol / L is a unit of molar concentration. These are the number of moles of dissolved material per liter of solution. 1 mol / L is also called 1M or 1molar. Mol / m3 is also a unit of molar concentration.
Molarity is expressed in units of moles per liter (mol / L). This is a very common unit, so it has its own symbol, which is the uppercase M. A solution with a concentration of 5 mmol / l is called a 5 M solution or has a concentration value of 5 mol.
The molar concentration of the solution is equal to the number of moles of the solute divided by the mass of the solvent (kilogram), and the molar concentration of the solution is equal to the number of moles of the solute divided by the volume of the solution (liter). increase.
Which of the following represents beta decay
OA. Tc-TC+y
O B.
B. 14Gd→ 144Sm+ He
O C. 160Eu+e→ 169 Sm
62
O D.
D.
63
164Gd→ ¹6 Tb + e
160
65
The correct answer that represents beta decay is
D. 164Gd → 164Tb + e, What happens in beta decayIn beta decay, a neutron in the nucleus is converted into a proton, and an electron (or beta particle) and an antineutrino are emitted from the nucleus.
In this case, a neutron in the 164Gd nucleus is converted into a proton, and an electron is emitted from the nucleus, resulting in the production of 164Tb.
Option A is not a valid representation of any known type of radioactive decay.
Option B represents alpha decay, in which an alpha particle is emitted from the nucleus.
Option C represents electron capture, in which an electron is captured by the nucleus.
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What happens when a solid is dissolved into a liquid?
.