Answer:
Body-centered cubic
Explanation:
As the edge length of the cubic unit is 350pm, its volume is:
(350x10⁻¹²m)³ = 4.288x10⁻²⁹m³. In cm³
4.288x10⁻²⁹m³ ₓ ([100cm]³ / 1m³) = 4.288x10⁻²³cm³.
As the density of the structure is 0.535g/cm³, the mass of a cubic unir is:
4.288x10⁻²³cm³ ₓ (0.535g / cm³) = 2.29x10⁻²³g,
As atomic mass of Lithium is 6.941g/mol. Mass of 1 atom of Li is:
6.941g/mol ₓ 1mol / 6.022x10²³ atoms = 1.1526x10⁻²³ g / atoms.
Thus, atoms of Li in 2.29x10⁻²³g are:
2.29x10⁻²³g ₓ ( 1 atom / 1.1526x10⁻²³g) = 2 atoms in 1 cubic unit
In body-centered cubic unit cell, there are 2 atoms per cubic unit cell. Thus, structure of Li is:
Body-centered cubicWhat is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12
Answer:
pH = 3.39
Explanation:
The equilibrium in water of ascorbic acid (With its conjugate base) is:
H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)
Where the acidic dissociation constant is written as:
Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]
H₂O is not taken in the Ka expression because is a pure liquid.
As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:
[H₂C₆H₆O₆] = 2.5x10⁻³M - X
[HC₆H₆O₆⁻] = X
[H₃O⁺] = X
Replacing in the Ka expression:
7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]
1.975x10⁻⁷ - 7.9x10⁻⁵X = X²
0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷
Solving for X:
X = -0.00048566→ False solution, there is no negative concentrations
X = 0.00040666 → Right solution
As [H₃O⁺] = X, [H₃O⁺] = 0.00040666
pH is defined as -log [H₃O⁺];
pH = -log 0.00040666,
pH = 3.391. What is the net ionic equation for the reaction that occurs when aqueous solutions of AgNO3 and CaCl2 are mixed and a precipitate forms? A. Ca+2(aq) + NO3-(aq) Ca(NO3)2(aq) B. Ag2+(aq) + 2Cl-(aq) AgCl2(s) C. Cl−(aq) + Ag+(aq) ⟶ AgCl(s) D. None of the above because no reaction occurs
Answer:
C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]
Explanation:
In this question our options are:
A. [tex]Ca^+^2_(_a_q_)+NO_3^-_(_a_q_)->Ca(NO_3)_2_(_a_q_)[/tex]
B. [tex]Ag^2^+_(_a_q_)+2Cl^-_(_a_q_)->AgCl_2_(_s_)[/tex]
C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]
D. None of the above because no reaction occurs
We have to remember that the ions produced by [tex]AgNO_3[/tex] are:
[tex]Ag^+[/tex] and [tex]NO_3^-[/tex]
And the ions produced by [tex]CaCl_2[/tex] are:
[tex]Ca^+^2[/tex] and [tex]Cl^-[/tex]
Additionally, we will have a double displacement reaction so the compounds produce are:
[tex]AgCl[/tex] and [tex]Ca(NO_3)_2[/tex]
If we remember the solubility rules, all the nitrate salts are soluble and the salts made with silver are not soluble. With this in mind, we will have a solid-state for [tex]AgCl_(_s_)[/tex] and an aqueous state for [tex]Ca(NO_3)_2_(_a_q_)[/tex].
If this is true, the final answer can be B or C. The charge of Ag is +1 so the final answer is C.
I hope it helps!
Calculate the mass of sodium nitrate( NaNO3), in grams, required to produce 128 g of oxygen, using the following equation: 2NaNO3 --> 2NaNO2 O2
Answer:
680g of NaNO3.
Explanation:
The balanced equation for the reaction is given below:
2NaNO3 —> 2NaNO2 + O2
Next, we shall determine the mass of NaNO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:
Molar mass of NaNO3 = 23 + 14 + (16x3) = 85g/mol
Mass of NaNO3 from the balanced equation = 2 x 85 = 170g
Molar mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32g
From the balanced equation above,
170g of NaNO3 decomposed to produce 32g of O2.
Now, we can obtain the mass of NaNO3 needed to produce 128g of O2 as shown below:
From the balanced equation above,
170g of NaNO3 decomposed to produce 32g of O2.
Therefore, Xg of NaNO3 will decompose to produce 128g of O2 i.e
Xg of NaNO3 = (170 x 128)/32
Xg of NaNO3 = 680g
Therefore, 680g of NaNO3 are needed to produce 128g of O2.
g A 25.00 mL sample of 0.0988 M benzoic acid (HC7H5O2, Ka = 6.5 · 10-5) is a monoprotic acid that is titrated with 0.115 M sodium hydroxide. Calculate the pH at the equivalence point.
Answer:
pH at equivalence point is 8.47
Explanation:
Benzoic acid react with NaOH, thus:
HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺
You reach equivalence point when moles of the acid = moles of NaOH.
Moles of benzoic acid are:
0.025L ₓ (0.0988mol / L) = 0.00247 moles
To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:
0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.
Total volume is 0.0465L.
There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:
0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻
C₇H₅O₂⁻ is in equilibrium with water, thus:
C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)
Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:
Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]
The concentrations in equilibrium of the species are:
[HC₇H₅O₂] = X
[OH⁻] = X
[C₇H₅O₂⁻] = 0.0531M - X
Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate
Replacing in Kb expression:
1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]
1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]
8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²
8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0
Solving for X:
X = -2.858x10⁻⁶M → False solution, there is no negative concentrations
X = 2.858x10⁻⁶M → Right solution
As [OH⁻] = X
[OH⁻] = 2.858x10⁻⁶M
pOH is -log [OH⁻]
pOH = 5.54
pH = 14 - pOH
pH = 8.46
pH at equivalence point is 8.47Place ~0.8 g Fe(NO3)3·9 H2O into a 10 mL volumetric flask and dissolve to the volumetric line with 0.1 M HNO3(aq). (Make sure you pre-rinse the volumetric flask with the 0.1 M HNO3(aq) solution before making the solution. Also, 9 waters of hydration (·9H2O) are present in this chemical, and must be included in the formula weight calculation.) Record the exact mass of iron nitrate used and show the calculation for the solution concentration in your observations.
Before lab, create a plan for the dilution of the Fe(NO3)3(aq) solution created above with DI water. This plan should be clearly outlined in your ELN. Create enough 0.0020 M Fe(NO3)3
How do I go about solving for the dilution plan?
Answer:
You can take 1mL of your stock solution in a 100mL volumetric flask and complete to volume.
Explanation:
You need to create a 0.00200M solution of Fe(NO₃)₃. First, you have to obtain the concentration of the first solution you made. That is:
0.8g Fe(NO₃)₃.9H₂O × (1mol / 403.9972g) =
0.0020 moles of Fe(NO₃)₃.9H₂O = Moles of Fe(NO₃)₃
In 10mL = 0.010L:
0.0020 moles of Fe(NO₃)₃ / 0.010L = 0.20M Fe(NO₃)₃
This is the concentration of your stock solution, as you want to obtain a 0.0020M solution, you dilution factor must be:
0.20M / 0.0020M = 100
That means you need to dilute your stock solution 100 times.
You can make this dilution, for example,
taking 1mL of your stock solution in a 100mL volumetric flask completing to volume with the solvent, 0.1M HNO₃(aq).Multiply each half reaction by the correct number, in order to balance charges for the two half reactions. Mg--> mg2++2c Au++e--> Au Please i need the answer quick i couldnt find it anywhere on the web
Answer:
Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)
Explanation:
The key to balancing redox reaction equations is this; ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation. The both half reaction equations can now be combined to give the overall reaction equation.
For the redox reaction under consideration;
Oxidation half equation;
Mg(s) ------> Mg^2+(aq) + 2e
Reduction half equation;
2Ag^+(aq) +2e ----> 2Ag(s)
Overall balanced redox reaction equation;
Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)
The elements will lose of gain electrons as needed to have an electron configuration that matches a noble gas.
a) true
b) false
Answer:
true
Explanation:
This is true because elements aim to have a full octet of electrons in their outermost (also called valence) shell. Noble gases already have a full valence shell which is why the elements that are not noble gases aim to be like them.
Answer:
a) true
Explanation:
this is the answer coz elements aim to have a full octet of electrons in their outermost (also called valence) shell.
what is the polarity of black pepper
Answer:
Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.
Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.
It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.
In general, the solubility of a short chain carboxylic acid is greater than an ester with the same number of carbons. For example, butyric acid (four carbon atoms) is infinitely soluble in water, whereas ethyl acetate is only partially soluble in water (8 g/100 mL).
Select the statement that correctly explains why the water solubility of the carboxylic acid is greater than that of the ester.
a. Ethyl acetate has two -CH3 groups while butyric acid has only one.
b. The carboxylic acid is much more soluble because the-COOH group is on the end of the molecule and is therefore more accessible to H2O.
c. An ester group (-COO-) cannot form hydrogen bonds with water since there are no hydrogen atoms in the ester group.
d. Carboxylic acid groups (-COOH) can form hydrogen bonds with each other, but ester groups (Coo-) cannot.
e. A carboxylic ad roup CH) can form more hydrogen bonds with water than an ester group (-COO-).
Answer:
A carboxylic acid group (-COOH) can form more hydrogen bonds with water than an ester group (-COO-).
Explanation:
The carboxylic acid group (-COOH) is found in the carboxylic acids. This group is ultimately responsible for the solubility of carboxylic acids in water. It is worthy of note that the high boiling points of low molecular weight carboxylic acids is often because they are capable of intermolecular hydrogen bonding which leads to the dimerization of carboxylic acid.
The solubility of carboxylic acids decreases as the length of the alkyl chain increases. Hence, a long chain carboxylic acid is less soluble in water than shorter chain carboxylic acids.
Ester molecules can't form hydrogen bonds with each other but they do form weak hydrogen bonds with water. This leads to the solubility of low molecular weight esters. However, if a carboxylic acid and an ester posses the same length of alky chain, the carboxylic acid will form more hydrogen bonds and thus be more soluble in water than than a corresponding ester of the same chain length.
What did Rutherford discover that Thomson did not understand?
Answer: That most of an atom's mass was packed in a central nucleus
Answer:
That most of an atom's mass was packed in the central nucleus.
Explanation:
Rutherford disproved Thomson's model in 1911 with his gold foil experiment, in which he demonstrated that the atom has a high-mass nucleus.
How much heat is liberated at constant pressure when 1.41 g of potassium metal reacts with 6.52 mL of liquid iodine monochloride (d = 3.24 g/mL)? 2K(s) + ICl(l) → KCl(s) + KI(s)
Answer:
The correct answer is - 13.33 kJ of heat
Explanation:
To know which one is the limiting reagent, determine the number of moles of each reagent in order .
n(K) = mass/atomic weight = 1.41/39 = 0.036 moles
Density of ICl = Mass/Volume
3.24 = Mass/6.52
Mass of ICl = 21.12 g
n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles
2 moles of K reacts with 1 mole of ICl
0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl
since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:
2 moles of K gives out -740.71 kJ of heat
1 mole of K will give out = -740.71/2 = 370.36 kJ of heat
0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat
Thus, the correct answer is - 13.33 kJ of heat
The amount of heat liberated at constant pressure is -13.33 kJ
The given parameters are:
Mass of potassium metal = 1.41gAmount of liquid iodine monochloride = 6.52 mLStart by calculating the number (n) of moles of each reagent using:
[tex]n = \frac{Mass}{Atomic\ weight }[/tex]
For the potassium metal, we have:
[tex]n_k = \frac{1.41g}{39g/mole}[/tex] ---where 39 is the atomic weight of potassium
[tex]n_k = 0.036\ moles[/tex]
For the liquid iodine monochloride, we start by calculating its mass using:
[tex]Mass = Density \times Volume[/tex]
So, we have:
[tex]Mass = 3.24 \times 6.52[/tex]
[tex]Mass = 21.12g[/tex]
The number of moles is then calculated as:
[tex]n_I=\frac{21.12g}{162.35}[/tex]
[tex]n_I = 0.130\ moles[/tex]
The reaction equation 2K(s) + ICl(l) → KCl(s) + KI(s) means that:
2 moles of potassium reacts with 1 mole of liquid iodine monochloride.
So, 0.036 moles of potassium will react with the following moles of liquid iodine monochloride.
[tex]A = \frac{0.036}{2}[/tex]
[tex]A = 0.018\ moles[/tex]
i.e. 0.036 moles of potassium will react with of liquid iodine monochloride
By comparison: 0.018 moles is less than 0.036 moles
So, the amount of heat liberated at constant pressure is:
[tex]Amount = 0.036 \times -\frac{740.71}{2} kJ[/tex]
[tex]Amount = -13.33 kJ[/tex]
Hence, the amount of heat liberated at constant pressure is -13.33 kJ
Read more about heat at:
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Identify the most and the least acidic compound in each of the following sets.
a. 2-chlorobutanoic acid:_______ 4-chlorobutanoic acid:_______ 3-chlorobutanoic acid:______.
b. 2,4-dinitrobenzoic acid:______ p-nitrobenzoic acid:______ p-bromobenzoic acid:_______.
c. p-cyanobenzoic acid:________ benzoic acid:_______ p-aminobenzoic acid:______.
Answer:
See explanation
Explanation:
Our answer options for this question are:
a. 2-chlorobutanoic acid:_______ 2-chlorobutanoic acid:_______ 3-chlorobutanoic acid:______.
b. 2,4-dinitrobenzoic acid:______ p-nitrobenzoic acid:______ p-bromobenzoic acid:_______.
c. p-cyanobenzoic acid:________ benzoic acid:_______ p-aminobenzoic acid:______
We have to check each set of molecules
a. 2-chlorobutanoic acid, 3-chlorobutanoic acid
In this case, the difference between these molecules is the position of "Cl". If the chlorine atom is closer to the acid group, we will have a higher inductive effect. So, the bond O-H would be weaker and we will have more acidity. So, the molecule with more acidity is 2-chlorobutanoic acid and the less acidic would be 3-chlorobutanoic acid.
b. 2,4-dinitrobenzoic acid, p-nitrobenzoic acid, p-bromobenzoic acid
In this case, we have several structural differences. In all the structure, we have deactivating groups ([tex]Br[/tex] and [tex]NO_2[/tex]). If we have a deactivating group the acidity will increase. In the case of "Br", we have a weak deactivating, so, this will be the less acidic one (p-bromobenzoic acid)
in 2,4-dinitrobenzoic acid we have two deactivating groups, therefore, this would be the most acid compound.
c. p-cyanobenzoic acid, benzoic acid, p-aminobenzoic acid
On these molecules, we have several structural differences. In p-cyanobenzoic acid we have a deactivating group, therefore in this molecule we will have more acidity. In the p-aminobenzoic acid, we have an activating group, so, this would be the less acidic compound.
See figure 1
I hope it helps!
Consider the following reaction where Kc = 6.50×10-3 at 298 K: 2NOBr(g) 2NO(g) + Br2(g) A reaction mixture was found to contain 9.83×10-2 moles of NOBr(g), 5.44×10-2 moles of NO(g), and 4.13×10-2 moles of Br2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc
Answer:
This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.
For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].
Explanation:
A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].
Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:
[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].
Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].
(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)
Calculate the reaction quotient given the concentration of each species:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].
(Note that the unit is ignored.)
Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,
Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:
reduce the value of the numerator, increase the value of the denominator, orboth.In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.
What happens to electrons in the photoelectric effect?
A. They are added to atoms.
B. They are knocked away from atoms.
C. They are turned into light.
D. They are made to orbit faster.
The correct answer is B. on Apex!
They have been knocked away from atoms. Hence, option B is correct.
What is the photoelectric effect?The photoelectric effect is a phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect.
This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
Hence, option B is correct.
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7. An element's most stable ion forms an ionic compound with chlorine having the formula XCl2. If the ion of element X has a mass of 89 and 36 electrons, what is the identity of the element, and how many neutrons does it have
Answer:
The element is strontium and the number of neutrons it have is 51.
Explanation:
Based on the given information, the ionic compound is,
XCl₂ ⇔ X₂⁺ + 2Cl⁻
X2+ is the ion of the mentioned element
As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,
No. of electrons = atomic number - charge
36 = atomic number - 2
Atomic number = 38
Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.
Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,
No. of neutrons = mass number - atomic number
= 89 - 38
= 51 neutrons.
2. Using Tables 5 and 6 below, balance the following chemical reaction:
Na2CO3 (aq)+CaCl2 (aq) → CaCO3 (s)+NaCl (aq)
Answer:
Na₂CO₃ + CaCl₂ → CaCO₃ + 2 NaCl
Explanation:
Na₂CO₃ + CaCl₂ → CaCO₃ + NaCl
First, determine what is balanced.
Ca and CO₃ is balanced; there is one on each side.
Na and Cl aren't balanced. On one side, there is two of each, while on the other there is only one. To fix this add a 2 in front of NaCl. This will cause the equation to be balanced.
Na₂CO₃ + CaCl₂ → CaCO₃ + 2 NaCl
What volume of water is required to dilute 120 cm3 of 10 mol dm–3 sulphuric acid to a concentration of 2 mol dm–3?
Answer:
0.48 dm3 (or 480 cm3)
Explanation:
First find the original no. of moles existing in the sulphuric acid:
no. of moles = volume (in dm3) x concentration
= 120/1000 x 10
= 1.2 mol
Then let the total volume of the diluted acid be v dm3.
Since
Concentration = no. of moles / volume,
so by substituting the given information,
2 = 1.2 / v
v = 0.6 dm3
Hence, the volume of water required
= 0.6 - 120/1000
= 0.48 dm3 (or 480 cm3)
Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].
First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.
Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.
In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.
A dilution is mathematically expressed as:
Ci×Vi = Cf×Vf
where
Ci: initial concentration Vi: initial volume Cf: final concentration Vf: final volume
In this case, you know:
Ci= 10 [tex]\frac{mol}{dm^{3} }[/tex] Vi= 120 cm³ Cf= 2 [tex]\frac{mol}{dm^{3} }[/tex] Vf= ?Replacing in the definition of dilution:
10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³= 2 [tex]\frac{mol}{dm^{3} }[/tex]× Vf
Solving:
Vf= (10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³) ÷2 [tex]\frac{mol}{dm^{3} }[/tex]
Vf= 600 cm³
In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].
Learn more about dilution:
brainly.com/question/20113402?referrer=searchResults brainly.com/question/22762236?referrer=searchResultsThe NaOH solution is standardized (or its true concentration) is found by reacting it with KHSO4. One of the two products from when NaOH reacts with KHSO4 is H2O. The other product is is a salt consisting of what?
a. NaK (aq)
b. (aq)
c. NaS (aq)
d. None of the above
Based on the Valence Shell Electron Pair Repulsion Theory (or VSEPR), molecules will arrange to keep the following as far apart as physically possible
a) mobile electrons
b) valence electron pairs
c) inner shell electrons
d) the electrons closest to the nuclei
Answer:
B. Valence Electron Pairs
Explanation:
Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.
What is the specific heat of a metal with a mass of 14.0 g, heat of 3.45 kJ and a change in temperature of 3.2 ℃?
determine the rate of reaction that follows the rate= k[A]^m[B]^n
rate=0.2*3^1*3^2=0.2*3*9=5.4(mol/L)s so the correct answer is C.
Why does the excess of base used in these eliminations favor the E2 over the E1 mechanism for elimination
Answer:
The base is involved in the rate determining step of an E2 reaction mechanism
Explanation:
Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;
Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.
For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;
Rate = k[alkyl halide] [base]
The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.
Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.
3A 2B --> 5C If compound A has a molar mass of 159.7 g/mol and compound C has a molar mass of 57.6 g/mole, how many grams of compound C will be produced from 18.24 grams of compound A and excess compound B
Answer:
10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.
Explanation:
3A + 2B ⇒ 5C
By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) the following amounts of reagent and products participate in the reaction:
A: 3 molesB: 2 molesC: 5 molesThe excess reagent will be that which is not completely depleted during the reaction.
The amount of product obtained from the reaction will always depend on the amount of limiting reagent in the reaction. Then, being B the excess reagent and therefore A the limiting reagent and knowing that compound A has a molar mass of 159.7 g/mole and compound C has a molar mass of 57.6 g/mole, by stoichiometry the following mass amounts of A and C participate in the reaction:
A: 3 moles* 159.7 g/mole= 479.1 gC: 5 moles* 57.6 g/mole= 288 gThen it is possible to apply the following rule of three: if by stoichiometry of the reaction 479.1 grams of A produce 288 grams of C, 18.24 grams of A, how much mass of C does it produce?
[tex]mass of C=\frac{18.24 grams of A*288 grams of C}{479.1 grams of A}[/tex]
mass of C= 10.96 grams
10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.
Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light?
1. Gamma rays
2. Infrared radiation
3. Ultraviolet liht
4. X-rays
5. Microwaves
6. Radio waves
Answer:
3,4,1 and 6,5,2
Explanation:
In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;
Radio waves
Microwaves
Infrared waves
Visible light rays
Ultraviolet rays
X-rays
Gamma rays
(a simple mnemonic is RMIVUXG)
The solubility of O2 in water is approximately 0.00380 g L-1 of water when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 760. torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?
Answer:
The correct answer is 0.00080 gram per liter.
Explanation:
Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.
Based on Henry's law,
Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas
Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas
= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr
= 0.00012 mol/L/atm
= 0.00012 M/atm
Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure
= 0.210 × 1 atm
= 0.210 atm
Now putting the values in Henry's law equation we get,
Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,
= 0.000025 mol/L × 32 gram/mol
= 0.00080 gram per liter
How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water
Answer:
WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.
Explanation:
Equation for the reaction:
K2SO4 + H20 ------->2 K+ + SO4^2-
When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.
1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-
At STP, 1 mole of K2SO4 will be the molar mass of the substance
Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol
Molar mass = 174 g/mol
So therefore;
1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion
When 63.7 g is used; we have:
174 g = 2 moles of K+
63.7 g = ( 63.7 * 2 / 174) moles of K+
= 0.73 moles of K+
Forr sulfate ion, we have:
174 g = 1 mole ofSO4^2-
63.7 g = (63.7 * 1 / 174) moles of SO4^2-
= 0.366 moles of SO4^2-
In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.
A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the
ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.
Answer:
The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.
Explanation:
Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.
The mass of the precipitate given is 22.73 grams.
Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams
Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.
And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.
Now the equation will become,
3x/78 + 45.46-2x/92.94 = 0.647 moles
0.03846 x + 0.489 - 0.02152 x = 0.647
0.01694 x + 0.489 = 0.647
0.01694 x = 0.158
x = 0.158/0.01694
x = 9.327 grams
Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,
= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams
Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams
the mass of Co²⁺ will be,
= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams
Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16 0 mL of CO
2 CO(g) + O2(g) 4, 2 CO2 (g)
Express your answer with the appropriate units.
Answer:
[tex]V_{CO_2}=16.0mL[/tex]
Explanation:
Hello,
In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:
[tex]n_{CO}V_{CO}=n_{CO_2}V_{CO_2}[/tex]
Thus, solving for the yielded volume of carbon dioxide we obtain:
[tex]V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL[/tex]
Best regards.
Write the following isotope in nuclide notation: oxygen-14
Answer:
[tex]14\\8[/tex]O
Explanation:
The top number always represents the mass number.
The bottom number always represents the atomic number.
The element always goes after the numbers.
If charge is present, that comes after the element.
How many moles of chloride ions are there in 2.5 L of 5 M magnesium chloride?
Answer:
[tex]n_{Cl^-}=25molCl^-[/tex]
Explanation:
Hello,
In this case, since the given 5-M concentration of magnesium chloride is expressed as:
[tex]5\frac{molMgCl_2}{L}[/tex]
We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:
[tex]n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-[/tex]
Best regards.