A compound turns completely into a liquid this observation best describes the physical appearance of a compound when it reaches the end of its melting point range. Here option B is the correct answer.
When a solid compound is heated, it undergoes a process called melting in which it transforms into a liquid state. The melting point of a compound is the temperature at which it changes from a solid to a liquid state. The melting process is characterized by a range of temperatures over which the compound is observed to be partially or fully melted.
The observation that best describes the physical appearance of a compound when the end of its melting point range is reached is B - the compound completely converts to a liquid. At the end of the melting point range, the compound has absorbed enough heat energy to fully overcome the intermolecular forces that hold its constituent particles together in a solid state, resulting in the complete transformation of the compound into a liquid.
This state is characterized by the loss of a crystalline structure, where the particles are free to move about and slide past each other, leading to an increased fluidity and mobility of the compound. At this stage, the compound is fully melted and can be poured or transferred into a new container in its liquid form.
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Complete question:
Which observation best describes the physical appearance of a compound when the end of its melting point range is reached?
A - the compound begins to convert to a liquid.
B - the compound completely converts to a liquid.
C - the compound begins to evaporate.
how many moles of naf must be dissolved in 1.00 liter of a saturated solution of pbf2 at 25˚c to reduce the [pb2 ] to 1 x 10–6 molar? (ksp pbf2 at 25˚c = 4.0 x 10–8)
The moles of NaF that must be dissolved in 1.00 liter of a saturated solution of PbF₂ at 25˚C to reduce the [Pb²⁺] to 1 x 10⁻⁶ molar is 2.0 x 10⁻⁵.
The solubility product expression for PbF₂ is given by:
Ksp = [Pb²⁻][F-]²At equilibrium, the product of the ion concentrations must be equal to the solubility product constant. We are given that the [Pb²⁺] in the saturated solution is 1 x 10⁻⁶ M. Therefore, we can use the Ksp expression to calculate the concentration of F- in the solution:
Ksp = [Pb²⁺][F⁻]²4.0 x 10⁻⁸ = (1 x 10⁻⁶)([F⁻]²)[F⁻]² = 4.0 x 10⁻²[F⁻] = 2.0 x 10⁻¹Now, we can calculate the amount of NaF needed to reduce the [F⁻] concentration to 2.0 x 10⁻¹ M. Since NaF is a 1:1 electrolyte, the concentration of F- will be equal to the concentration of NaF added.
Number of moles of NaF = (2.0 x 10⁻¹) mol/L x 1.00 L = 2.0 x 10⁻¹ molesHowever, we need to dissolve this amount of NaF in a saturated solution of PbF₂. Therefore, we need to check that the amount of NaF we added will not exceed the maximum amount that can dissolve in the solution at 25˚C.
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The cloud droplets in a cloud are formed by water vapor molecules and: A) protons. B) ions. C) molecules of air. D) condensation nuclei.
Answer:
condensation nuclei
Explanation:
PLEASE ANSWER ASAP
1. How many atoms are present in 8.500 mole of chlorine atoms?
2. Determine the mass (g) of 15.50 mole of oxygen.
3. Determine the number of moles of helium in 1.953 x 108 g of helium.
4. Calculate the number of atoms in 147.82 g of sulfur.
5. Determine the molar mass of Co.
6. Determine the formula mass of Ca3(PO4)2.
IT WOULD BE HELPFUL
1) 5.1167 x 10²⁴atoms of chlorine. 2) 248.00 g. 3) 4.8825 x 10⁷ moles of helium. 4) 2.7757 x 10²⁴ atoms of sulfur. 5) Molar mass of Co (cobalt) is 58.93 g/mol. 6) Formula mass = 310.18 g/mol.
What is meant by formula mass?Sum of the atomic masses of all the atoms in chemical formula is called formula mass
1.) Number of atoms = 8.500 moles x 6.022 x 10²³ atoms/mole = 5.1167 x 10²⁴ atoms of chlorine.
2.) Molar mass of oxygen is 16.00 g/mol. Therefore:
Mass of 15.50 moles of oxygen = 15.50 moles x 16.00 g/mol = 248.00 g.
3.) Molar mass of helium is 4.00 g/mol. Therefore, the number of moles of helium in 1.953 x 10⁸ g is:
Number of moles = 1.953 x 10⁸ g / 4.00 g/mol = 4.8825 x 10⁷ moles of helium.
4.) Molar mass of sulfur is 32.06 g/mol. Therefore, the number of moles of sulfur in 147.82 g is:
Number of moles = 147.82 g / 32.06 g/mol = 4.6084 moles of sulfur.
To find the number of atoms, we can use Avogadro's number again:
Number of atoms = 4.6084 moles x 6.022 x 10²³ atoms/mole = 2.7757 x 10²⁴ atoms of sulfur.
5.) Molar mass of Co (cobalt) is 58.93 g/mol.
6.) Ca₃(PO₄)₂ contains 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms.
Atomic masses of these elements are:
Calcium (Ca) = 40.08 g/mol
Phosphorus (P) = 30.97 g/mol
Oxygen (O) = 16.00 g/mol
Therefore, formula mass of Ca₃(PO₄)₂ is:
Formula mass = (3 x 40.08 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)
= 120.24 g/mol + 61.94 g/mol + 128.00 g/mol
= 310.18 g/mol.
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determine the standard enthalpy change for the decomposition of hydrogen peroxide per mole of hydrogen peroxide.
The standard enthalpy change for the decomposition of hydrogen peroxide per mole of hydrogen peroxide is -98.2 kJ/mol.
when 1 mole of hydrogen peroxide (H2O2) ( H 2 O 2 ) undergoes decomposition, the heat evolved (ΔH) is −98.2kJ. − 98.2 k J . The molar mass of H2O2 H 2 O 2 is 34.015 g/mol. This means that the mass of 1 mole of H2O2 H 2 O 2 is 34.015 g.
This value is obtained from the standard enthalpy of formation of the products (H2 and O2) and the standard enthalpy of formation of the reactant (H2O2). Enthalpy of formation is the energy change that occurs when a compound is formed from its elements, in their standard states.
The difference between the enthalpies of formation of the products and the reactant is the enthalpy change for the reaction. In this case, the enthalpy change for the decomposition of hydrogen peroxide is -98.2 kJ/mol. This indicates that the decomposition of hydrogen peroxide is an exothermic reaction and it releases 98.2 kJ/mole of energy.
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What is the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K?
Answer:
0.9g/L.
Explanation:
To calculate the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K, we can use the ideal gas law:
PV = nRT
where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles of gas, R is the universal gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin (K).
We can rearrange this equation to solve for the number of moles of gas:
n = PV / RT
Next, we can use the molar mass of H2S (34.08 g/mol) to convert the number of moles to mass:
mass = n × molar mass
Finally, we can divide the mass by the volume to obtain the density:
density = mass/volume
Let's assume a volume of 1 L (since the volume is not given in the question). Then we have:
P = 0.7 atm
T = 322 K
R = 0.08206 L·atm/(mol·K)
molar mass of H2S = 34.08 g/mol
First, we calculate the number of moles of H2S using the ideal gas law:
n = PV / RT
n = (0.7 atm) (1 L) / (0.08206 L·atm/(mol·K) × 322 K)
n = 0.0265 mol
Next, we calculate the mass of H2S using the number of moles and the molar mass:
mass = n × molar mass
mass = 0.0265 mol × 34.08 g/mol
mass = 0.9 g
Finally, we calculate the density of H2S:
density = mass/volume
density = 0.9g/1 L
density = 0.9 g/L
Therefore, the density of hydrogen sulfide (H2S) at 0.7 atm and 322 K is approximately 0.9g/L.
a 1.25 g sample of co2 is contained in a 750. ml flask at 22.5 c. what is the pressure of the gas, in atm?
The pressure of gas is 1.05 atm when a 1.25 g sample of CO₂ is contained in a 750ml flask at 22.5°C.
Molecular weight of CO₂ is 1.25g ,Volume of CO₂ is 750ml,Temperature of CO₂ is 22.5°C and the gas constant is 0.08206 L atm/mol K.
Using the ideal gas law equation the pressure is found to be 1.05 atm.
To calculate the pressure of the gas, we can use the ideal gas law equation: [tex]PV=nRT[/tex]
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the volume to liters by dividing by 1000: 750 ml = 0.75 L.
Next, we need to calculate the number of moles of CO₂ present in the flask. We can use the molecular weight of CO₂ to convert from grams to moles:
[tex]1.25 * (1 /44.01 ) = 0.0284 mol[/tex]
Now we can plug in the values into the ideal gas law equation:
[tex]PV=nRT[/tex]
[tex]P * 0.75 L = 0.0284 mol * 0.08206 L*atm/mol*K * (22.5 + 273.15) K[/tex]
Simplifying and solving for P, we get:
[tex]P = (0.0284 * 0.08206 * 295.65) / 0.75 = 1.05 atm[/tex]
Therefore, the pressure of the gas in the flask is 1.05 atm.
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the molar solubility of pbi 2 is 1.5 × 10 −3 m. calculate the value of ksp for pbi 2 .4.5 x 10 -6
The value of Ksp for PbI2 is 4.05 × 10^-8 if the molar solubility of PBI 2 is 1.5 × 10 −3 m.
The molar solubility of PBI 2 = 1.5 × 10 −3 m
The solubility product constant = 2 .4.5 x 10 -6
The solubility product constant (Ksp) for PbI2 can be estimated using the molar solubility of PbI2, the stoichiometry of the equilibrium equation is:
[tex]PbI2(s) = Pb2+(aq) + 2I-(aq)[/tex]
The equation for Ksp is:
Ksp = [tex][Pb2+][I-]^2[/tex]
[Pb2+] = S = 1.5 × 10−3 M,
[I-] = 2S = 3 × 10−3 M
The stoichiometric coefficient of I- is 2. Substituting these values into the Ksp equation we get:
Ksp =[tex](1.5 × 10^-3) × (3 × 10^-3)^2[/tex]
Ksp = 4.05 × 10^-8
Therefore, we can conclude that the value of Ksp for PbI2 is 4.05 × 10^-8.
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The value of Ksp for PbI2 is 3.375 × 10^-9 or 4.5 x 10 -6. The expression for the solubility product constant (Ksp) of a sparingly soluble salt such as PbI2 is: Ksp = [Pb2+][I-]^2
where [Pb2+] and [I-] are the molar concentrations of the lead ion and iodide ion, respectively, in a saturated solution of PbI2.
Given that the molar solubility of PbI2 is 1.5 × 10^-3 M, we can assume that [Pb2+] and [I-] in the saturated solution are also equal to 1.5 × 10^-3 M. Therefore, we can substitute these values into the Ksp expression and solve for Ksp:
Ksp = (1.5 × 10^-3 M)(1.5 × 10^-3 M)^2
Ksp = 3.375 × 10^-9
So the value of Ksp for PbI2 is 3.375 × 10^-9 or 4.5 x 10 -6 (if that was a typo in the question).
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What is the concentration (in molality) of an aqueous solution of NaCl made by adding
4.56 g of NaCl to enough water to give 20.0 mL of solution. Assume the density of the
solution is 1.03 g/mL
Answer:
data given
mass of NaCl 4.56
dissolved volume 20ml(0.02l)
density of solution 1.03g/ml
Required molality
Explanation:
molarity=m/mr×v
where
m is mass
mr molar mass
v is volume
now,
molarity=4.56/58.5×0.02
molarity =3.9
: .molarity is 3.9mol/dm^3
According to molal concentration, the concentration (in molality) of an aqueous solution of NaCl is 0.0047 mole/kg.
What is molal concentration?Molal concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molal concentration is moles/kg.
The molal concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molal concentration is calculated by the formula, molal concentration=mass/ molar mass ×1/mass of solvent in kg.
In terms of moles, it's formula is given as molal concentration= number of moles /mass of solvent in kg.
Substitution in formula gives the answer but first mass of solution is determined which is density×volume= 1.03×20=20.6 g , mass of solvent= 20.6-4.56=16.05, thus molal concentration=4.56/58.5×1/16.05=0.0047 moles/kg.
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consider the reaction performed in the sn1 lab. what would be the effect on the rate of the reaction if 2-propanol (isopropanol) was used instead of 2-methyl-2-propanol (t-butanol) assuming only an sn1 reaction occurs? group of answer choices the rate of the reaction would decrease, because the secondary carbocation is more difficult to form. the rate of the reaction would increase, because the secondary carbocation is easier to form. there would be no difference in reaction rate. the reaction would not proceed at all.
The rate of the reaction is directly proportional to the stability of the carbocation intermediate, and any changes in the solvent will affect the rate of the reaction.
In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate. The stability of the carbocation intermediate affects the rate of the reaction.
In this case, if 2-propanol (isopropanol) was used instead of 2-methyl-2-propanol (t-butanol), the rate of the reaction would decrease. This is because the carbocation intermediate formed in 2-propanol is less stable compared to the one formed in t-butanol.
The carbocation intermediate formed in t-butanol is tertiary, which is more stable than the one formed in isopropanol, which is secondary. This means that the reaction will be slower in isopropanol due to the less stable carbocation intermediate.
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