The correct answer is C. Surface craters
Explanation:
Short-term environmental changes involve temporary changes and effects in the ecosystem, which are mainly minor. In the case of a small asteroid or comet, this will likely lead to surface craters or changes in the surface of the impact zone. This is because the craters and asteroids impact the surface at hight speed. Also, because this is a minor event it might lead to the death of some organisms but not the extinction of these and it is not expected this has major effects such as changes in weather. Thus, the short-term effect that this will most likely cause is "surface craters."
Answer:
surface
Explanation:
what is the value of the equilibrium constant at 500k for a chemical equilivrium that has a delta h value of 250kj mol and s value of 48 j mol k
The basic function of a carburetor of an automobile is to atomize the gasoline and mix it with air to promote rapid combustion. As an example, assume that 30 cm3 of gasoline is atomized into N spherical droplets, each with a radius of 2.0 × 10−5 m. What is the total surface area of these N spherical droplets? Answer: [A] m2.
Answer:
The total surface area of these N spherical droplets is 4.4929 m²
Explanation:
From the information given :
assuming that :
30 cm³ of gasoline is atomized into N spherical droplets &
each with a radius of 2.0 × 10−5 m
We are tasked to determine the total surface area of these N spherical droplets
We all known that:
[tex]1 \ cm^3 = 10 ^{-6} m^3[/tex]
Therefore
[tex]30 \ cm^3 = 30 * 10 ^{-6} m^3 = 3 *1 0^{-5} \ m^3[/tex]
For each droplet; there is a required volume which is = [tex]\dfrac{4}{3} \pi r ^3[/tex] since it assumes a sphere shape .
Thus;
replacing radius(r) with 2.0 × 10−5 m; we have:
[tex]= \dfrac{4}{3} \pi * (2.0 *10^{-5} m) ^3[/tex]
= [tex]3.35 * 10^{-14} \ m^3[/tex]
However; there are [tex]3*10^{-5} \ m^3[/tex] gasoline atomized into N spherical droplets with each with radius 2.0 × 10−5 m
For N ; we have ;
[tex]=\dfrac{3*10^{-5} \ m^3}{3.35 * 10^{-14} \ m^3/ droplet}[/tex]
= [tex]8.95*10^8 \ droplet s[/tex]
So; each droplet have a surface area = [tex]4 \pi r^2[/tex]
= [tex]4 \pi (2.0*10^{-5}m) ^2[/tex]
= [tex]5.02*10^{-9} \ m^2/droplets[/tex]
The surface area per droplet is equivalent to [tex]5.02*10^{-9} \ m^2/droplets[/tex]
Thus;
The total surface area of these N spherical droplets will be :
= [tex]8.95*10^8 \ droplet s * 5.02*10^{-9} \ m^2/ droplets[/tex]
= 4.4929 m²
The total surface area of these N spherical droplets is 4.4929 m²
At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.00317sâ1: 2N2O5(g) â2N2O4(g) + O29(g) Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.
Answer:
[A] = 1.438M = 1.4M (Two s.f)
Explanation:
Rate constant, k = 0.00317
Initial Concentration, [A]o = 1.44M
Final Concentration, [A] = ?
Time, t = 0.240 s
Since this is a second order reaction, the formula for this is given as;
1 / [A] = 1 / [A]o + kt
1 / [A] = 1 / 1.44 + (0.00317 * 0.240)
1 / [A] = 0.6944 + 0.0007608
1 / [A] = 0.6952
[A] = 1.438M = 1.4M (Two s.f)
A gas contained in a steel tank has a volume of 1.5 L at a temperature of 390 K. What will be the volume when the temperature changes to 1470C? Group of answer choices
Answer:
1.5 L
Explanation:
If the gas is contained in a steel tank, the volume will remain constant when the temperature changes.
The volume will be 1.5 L.
The force that opposes drag and is powered by combustion reactions in the
engine is
Answer:
Thrust.
Explanation:
hope this helps you :)
Answer:
thrust
Explanation:
If 50 ml of 1.00 M of H2SO4 and 50 ml of 2.0 M KOH are mixed what is the concentration of the resulting solutes?
Answer: [H2SO4] = 0.5M;
[KOH] = 1M
Explanation: Molarity is the solution concentration defined by:
molarity = [tex]\frac{mol}{L}[/tex] or M
To determine the concentration of the mixture, find how many mols of each compound there are in the mixture:
50 mL = 0.05L
H2SO4
1 mol/L * 0.05L = 0.05mol
KOH
2mol/L * 0.05L = 0.1 mol
The mixture has a total volume of:
V = 50 + 50 = 100 mL = 0.1 L
The concentration of the resullting solutes:
[H2SO4] = [tex]\frac{0.05}{0.1}[/tex] = 0.5 M
[KOH] = [tex]\frac{0.1}{0.1}[/tex] = 1 M
Concentration of H2SO4 is 0.5M while for KOH is 1M.
Draw structural formulas for the major organic product(s) of the reaction shown below.
• You do not have to consider stereochemistry.
If no reaction occurs, draw the organic starting material.
Remember to include all of the formal charges on the atoms of any nitro groups.
Answer:
3-bromobenzoic acid
Explanation:
In this case, we have to remember that the [tex]Br_2/FeBr_3[/tex] is a reaction in which we add Br into the molecule an electrophilic aromatic substitution. Additionally, we have a carboxylic acid in the benzene. This carboxylic acid is an ortho director because is a deactivating group (it removes electrons from the benzene ring). With this in mind, a "Br" atom would be added in an ortho position respect to the COOH group and we will obtain 3-bromobenzoic acid.
See figure 1.
I hope it helps!
To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group according to electrophilic aromatic substitution.
Electrophilic aromatic substitution is a type of organic reaction in which an atom or group in an aromatic ring is substituted with an electrophile. It is a fundamental reaction in aromatic chemistry that happens due to the aromatic system's high electron density.
It is an electrophilic aromatic substitution process in which Br is incorporated into the molecule. In addition, the benzene contains a carboxylic acid. Because it removes electrons from the benzene ring, this carboxylic acid functions as an ortho director. To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group. The product is seen in the photographs below.
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Which of the following would be useful for converting g/mol to g/L?
A. Mass percent
B. Avogadro's number
C. Molarity
D. Molar mass
Answer:
Molarity
Explanation:
The conversion of g/mol to g/L molarity can be used. Thus, option C is correct.
The g/mol has been the amount of solute present in a mole. The g/mol has been the molecular weight of the compound.
The g/L has been the mass of solute present in a L of solution. The g/L has the unit for density.
Molarity has been the moles of solute present in the liter of solution. It has been given as mol/L.
The product of g/mol and g/L gives the value of mol/L. Thus, to convert g/mol to g/L molarity can be used. Thus, option C is correct.
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A compound consisting of atoms of small atomic mass is more likely to require what
Answer:
a lower temperature to liquefy
Explanation:
What is the molarity of a solution that is 7.00% by mass magnesium sulfate and has a density of 1.071 g/mL?
Answer:
0.623 M
Explanation:
Step 1: Given data
Percent by mass (%m/m): 7.00 %Density of the solution (ρ): 1.071 g/mLMolar mass of magnesium sulfate: 120.37 g/molStep 2: Calculate the percent by volume (%m/v)
We will use the following expression.
[tex]\%m/v = \%m/m \times \rho = 7.00\% \times 1.071g/mL = 7.50g\%mL[/tex]
Step 3: Calculate the molarity
7.50 g of magnesium sulfate are dissolved in 100 mL of the solution. The molarity is:
[tex]M = \frac{7.50g}{120.37g/mol \times 0.100L } = 0.623 M[/tex]
Write the empirical formula
Answer:
[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]
Explanation:
[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]
Identify the precipitation reaction in the set?
Answer:
The third reaction
(2NaOH + NiCL2 ---> 2NaCl + Ni(OH)2)
Explanation:
By definition, a precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined.
(Source: lumenlearning)
From the 4 options, we can eliminate the first and second one immediately because there is no formation of an insoluble salt.
Then, the last one can also be eliminated because even though there is insoluble solid formed, but it is not a salt, and, the reactants are not solutions too. In fact, the last one is a displacement reaction. A more reactive metal displaces a less reactive metal to form an ion.
Since the third reaction matches the definition of precipitation reaction, this is the answer.
Use the reaction I2(s) I2(g), H = 62.4 kJ/mol, S = 0.145 kJ/(molK)
At what temperature is the reaction at equilibrium?
A.157K
B.430K
C.0.002K
D.62K
Answer: B. 430 K
Explanation:
According to Gibb's equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change = +62.4 kJ/mol
[tex]\Delta S[/tex] = entropy change = +0.145 kJ/molK
T = temperature in Kelvin
[tex]\Delta G[/tex] = +ve, reaction is non spontaneous
[tex]\Delta G[/tex] = -ve, reaction is spontaneous
[tex]\Delta G[/tex] = 0, reaction is in equilibrium
[tex]\Delta H-T\Delta S=0[/tex] for reaction to be spontaneous
[tex]T=\frac{\Delta H}{\Delta S}[/tex]
[tex]T=\frac{62.4kJ/mol}{0.145kJ/molK}=430K[/tex]
Thus the Reaction is spontaneous when temperature is 430 K.
Answer:
430 K
Explanation:
i just took the test on a pex :)
At a temperature of 393 K, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at a temperature of 478 K? (Assume constant volume)
Answer:
1.30atm
Explanation:
P1/T1 = P2/T2
1.07/393 = P2/478
Answer: the first one is correct
Explanation:
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How many water molecules are in a block of ice containing 1.25 mol of water (H2O)
Answer:
Molecules = 7.5 × 10²³ molecules
Explanation:
Given:
Moles = 1.25 mol
Avogadro's No. = [tex]N_{A}[/tex] = 6.022 * 10²³
Required:
Molecules = ?
Formula:
Molecules = Moles × [tex]N_{A}[/tex]
Solution:
Molecules = 1.25 × 6.022 × 10²³
Molecules = 7.5 × 10²³ molecules
If an electromagnetic wave has a frequency of 4.5 x 10^18 Hz, what is its wavelength? The speed of light is 3 x 108 m/s.
Answer:
Wavelength, λ = 6.7 x 10^-11 m
Explanation:
Frequency and wavelength are inversely proportional to each other.
In this problem;
f = 4.5 x 10^18 Hz
wavelength, λ = ?
Speed of light, c = 3 x 108 m/s.
These variables are related by the following equation;
c = λ * f
Making λ subject of focus, we have;
λ = c / f
λ = 3 x 10^8 / 4.5 x 10^18
λ = 0.67 x 10^-10
λ = 6.7 x 10^-11 m
What energy transfer happens when wood is burning?
Answer:
Mechanical to Heat
explanation:
The wood itself can make mechanical energy but when it's on fire it makes heat energy
Answer: Chemical to heat and light
Explanation: The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs and produces heat and light.
Morphine, C 17H 19NO 3, is often used to control severe post-operative pain. What is the pH of the solution made by dissolving 25.0 mg of morphine in 100. mL of water? (For morphine, K b = 1.62 × 10 –6.)
Answer:
pH = 9.58
Explanation:
First of all, we need to determine the molarity of the solution.
We determine the molar mass of morphine:
12g/m . 17 + 1 g/m . 19 + 14 g/m + 16 g/m . 3 = 285.34 g/m
molar mass g/m, is the same as mg/mm
25 mg . 1 mmol / 285.34 mg = 0.0876 mmoles / 100 mL = 8.76×10⁻⁴ M
In diltuted solution, we must consider water.
Mass balance for morphine = [Morphine] + [Protonated Morphine]
8.76×10⁻⁴ M = [Morphine] + [Protonated Morphine]
As Kb is too small, I can skipped, the [Protonated Morphine]
8.76×10⁻⁴ M = [Morphine]
In the charge balance I will have:
[OH⁻] = [H⁺ morphine] + [H⁺]
Let's go to the Kb expression
Morphine + H₂O ⇄ MorphineH⁺ + OH⁻ Kb
Kb = [MorphineH⁺] [OH⁻] / [Morphine]
Kb = [MorphineH⁺] [OH⁻] / 8.76×10⁻⁴ M
So now, we need to clear [MorphineH⁺] to replace it in the charge balance
Kb . 8.76×10⁻⁴ M / [OH⁻] = [MorphineH⁺]
Now, the only unknown value is the [OH⁻]
[OH⁻] = Kb . 8.76×10⁻⁴ M / [OH⁻] + Kw/[OH⁻]
Remember that Kw = [H⁺] . [OH⁻]
[H⁺] = Kw/[OH⁻]
[OH⁻]² = 1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴
[OH⁻] = √(1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴)
[OH⁻] = 3.76×10⁻⁵ → - log [OH⁻] = pOH = 4.42
pH = 14 - pOH → 14 - 4.42 = 9.58
Resources Use the exothermic and endothermic interactive to classify the solution process for each solute. Exothermic solution process Endothermic solution process
KOH CaCl, NaCT NaOH NaNO, NH NO,
Answer:
Exothermic interractive are the following: NaOH, KOH, CaCl₂
Endothermic interactive are the following: NaCl, NH₄NO₃, NaNO₃
Explanation:
NaOH, KOH, and CaCl2 are exothermic reactants. NaCl, NH4NO3, and NaNO3 are endothermic interacting substances.
Endothermic reactions: what are they?Chemical processes that can release or absorb energy are referred to as endothermic. In endothermic reactions, more energy is used when bonds in the reactants are broken than is released when new bonds are formed in the products.
How do endothermic processes take place?When the temperature of the an isolated system drops while the surroundings of the a non-isolated system warm up, this is known as an endothermic response. The heat of reaction is generally positive in endothermic processes (qrxn>0).
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Cual es la diferencia entre agua pesada y agua ligera a) el agua pesada contiene mas minerales que el agua ligera b) el agua ligera es liquida mientras el agua pesada es solida c) el agua ligera es agua purificada y el agua pesada es agua contaminada d) el agua pesada contiene mas elementos estearato de sodio
Answer:
d) El agua pesada contiene mas elementos
Explanation:
La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.
Answer:
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Explanation:ki
A sample of oxygen is collected over water at a total pressure of 692.2 mmHg at 17°C. The vapor pressure of water at 17°C is 14.5 mmHg. The partial pressure of the O2
Answer:
677.7 mmHg
Explanation:
The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.
Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.
Given that;
Total pressure of gas mixture = 692.2 mmHg
SVP of water at 17°C = 14.5 mmHg
Therefore, partial pressure of oxygen = 692.2-14.5
Partial pressure of oxygen = 677.7 mmHg
A solid white substance A is heated strongly in the absence of air. It decomposes to form a new white substance B and a gas C. The gas has exactly the same properties as the product obtained when carbon is burned in an excess of oxygen. Based on these observations, can we determine whether solids A and B and the gas C are elements or compounds?
Answer:
A, B and C are compounds
Explanation:
First of all, I need to establish that when carbon is burnt in excess oxygen, carbon dioxide is obtained as shown by this equation; C(s) + O2(g) ----> CO2(g).
Looking at the presentation in the question, A was said to be heated strongly and it decomposed to B and C. Only a compound can decompose when heated. Elements can not decompose on heating. Secondly, compounds usually decompose to give the same compounds that combined to form them. Compounds hardly decompose into their constituent elements.
Again from the information provided, the compound A is a white solid. This is likely to be CaCO3. It decomposes to give another white solid. This may be CaO and the gas was identified as CO2.
Hence;
CaCO3(s)--------> CaO(s) + CO2(g)
What is the product(s) of the reaction below?
2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)
A. Solid aluminum oxide and solid iron
B. Solid aluminum
C. Saturated aluminum oxide and saturated iron
D. Iron(III) oxide and aluminum oxide
Answer:
I would put A
Explanation:
A new substance is produced as a result of a chemical reaction in which bonds between the molecules of the reactant and product are broken and new bonds are formed. Here the products are Al₂O₃ and Fe. The correct option is A.
Chemical reactions are interactions between two or more molecules that result in the production of new products. Products, as opposed to reactants, are compounds that result from an interaction between two other substances.
The reactants are on the left, while the products that are created are on the right. A one-headed or two-headed arrow connects the reactants and products.
Thus the correct option is A.
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What are the correct formulas and coefficients for the products of the following double-replacement reaction? RbOH + H3PO4→
Answer:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Explanation:
Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
The lock and key model and the induced fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Indicate whether each statement is part of the lock and key model, the induced fit model, or is common to both models.
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions
Answer:
"The active site of the enzyme has a complementary rigid structure" belongs to the key and lock system
"The conformation of the enzyme changes when it binds to the substrate so that the active site conforms to the substrate." belongs to the induced fit system.
"The substrate binds to the enzyme at the active site, forming an enzyme-substrate complex" belongs to both, that is, the key and lock system and the induced fit system.
"The substrate binds to the enzyme through non-covalent interactions" can belong to both enzyme systems.
Explanation:
Enzymatic key and lock systems bear this name because the enzyme at its site of union with the substrate has an ideal shape so that its fit is perfect, similar to a headbreaker, so once they are joined they are not It can bind another substrate to the enzyme, since they are generally associated with strong chemical bonds.
The shape of the enzyme's active site is a negative of what the shape of the substrate would be.
On the other hand, in the mechanism or enzyme system of induced adjustment, the enzyme has an active site that is where it binds with the substrate and another site where another chemical component binds, which when this chemical component binds this enzyme changes its morphology and becomes "active" to bond with your substrate.
This happens a lot in the inactive enzymes that are usually activated in digestive processes since the fact that these enzymes are constantly active would be dangerous, therefore the body takes the induced enzyme system as a control mechanism, where a molecule or chemical compound induces change morphological of an enzyme by means of the allosteric union so that it joins its substrate and catalyzes or analyzes it, depending on the enzymatic character of the enzyme.
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and placed in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
a soluation of acetone in water has a molarity of 2.422M and a density of 0.970 g/mL. Calculate the mole fraction
Answer:
[tex]x_{acetone}=7.970x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given molarity, we can assume a volume of 1 L of solution, to obtain the following moles of acetone:
[tex]n=0.422mol/L*1L=0.422mol[/tex]
Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:
[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{0.970g}{1mL}=970g[/tex]
After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):
[tex]m_{H_2O}=970g-0.422molAcetone*\frac{58.08g\ Acetone}{1mol\ Acetone} =945.49gH_2O[/tex]
Next, the moles of water:
[tex]n_{H_2O}=945.49g*\frac{1molH_2O}{18gH_2O} =52.53molH_2O[/tex]
Finally, the mole fraction:
[tex]x_{acetone}=\frac{n_{acetone}}{n_{acetone}+n_{H_2O}}=\frac{0.422mol}{0.422mol+52.53mol}\\ \\x_{acetone}=7.970x10^{-3}[/tex]
Regards.
The mathematics of combining quantum theory with wave motion of atomic particles is known as _____.
Combining quantum theory with wave motion of atomic particles is: Wave Mechanics
A sample of magnesium ribbon is ignited in a crucible to form magnesium oxide. Determine the empirical formula of magnesium oxide from the following data:
mass of crucible and cover + magnesium metal
33.741 g
mass of crucible and cover
33.500 g
mass of crucible and cover + magnesium oxide
33.899 g
Answer:
MgO
Explanation:
The following data were obtained from the question:
mass of crucible and cover + magnesium metal = 33.741 g
mass of crucible and cover = 33.5 g
mass of crucible and cover + magnesium oxide = 33.899 g
Next, we shall determine the mass of magnesium metal. This can be obtained as follow:
mass of crucible and cover + magnesium metal = 33.741 g
mass of crucible and cover = 33.5 g
Mass of magnesium metal =..?
Mass of magnesium metal = (mass of crucible and cover + magnesium metal) – (mass of crucible and cover)
Mass of magnesium metal = 33.741 – 33.5
Mass of magnesium metal = 0.241g
Next, we shall determine the mass of magnesium oxide. This can be obtained as follow:
mass of crucible and cover + magnesium oxide = 33.899 g
mass of crucible and cover = 33.5 g
Mass of magnesium oxide =?
Mass of magnesium oxide = (mass of crucible and cover + magnesium oxide) – (mass of crucible and cover)
Mass of magnesium oxide = 33.899 –. 33.5
Mass of magnesium oxide = 0.399g
Next, we shall determine the mass of oxygen. This can be obtained as follow:
Mass of magnesium oxide = 0.399g
Mass of magnesium metal = 0.241g
Mass of oxygen =..?
Mass of oxygen = (Mass of magnesium oxide) – (Mass of magnesium metal)
Mass of oxygen = 0.399 – 0241
Mass of oxygen = 0.158g
Now, we can obtain the empirical formula for the magnesium oxide as follow:
Mg = 0.241g
O = 0.158g
Divide by their molar mass
Mg = 0.241 / 24 = 0.01
O = 0.158 / 16 = 0.0099
Divide by the smallest
Mg = 0.01 / 0.0099 = 1
O = 0.0099 / 0.0099 = 1
Therefore, the empirical formula for the magnesium oxide is MgO
A 3.00-g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 2.93 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (molar mass of PbSO4 = 303.3 g/mol)
Answer:
33.3% of Sn in the sample
Explanation:
The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.
Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:
2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.
As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:
9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺
As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g
And the percentage of Sn in the sample is:
1.00g / 3.00g ₓ 100 =
33.3% of Sn in the sample