Answer:12.9e-12g or in short 12.9pg
Explanation:as p=1e-12
what volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below
2co(g) + 2no(g) -> n2(g) + 2co2(g)
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
Answer:
37.8
Explanation:
A 1 liter solution contains 0.436 M hypochlorous acid and 0.581 M potassium hypochlorite. Addition of 0.479 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
Answer:
Exceed the buffer capacity and Raise the pH by several units
Explanation:
Options are:
Raise the pH slightly
Lower the pH slightly
Raise the pH by several units
Lower the pH by several units
Not change the pH
Exceed the buffer capacity
The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:
pH = pKa + log [ClO⁻] / [HClO]
pKa for hypochlorous acid is 7.53
pH = 7.53 + log [0.581M] / [0.436M]
pH = 7.65
Barium hydroxide reacts with HClO producing more ClO⁻, thus:
Ba(OH)₂ + 2HClO → 2ClO⁻ + 2H₂O
As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO
As you have just 0.436 moles (Volume = 1L),
The addition will:
Exceed the buffer capacityThe Ba(OH)₂ that reacts is:
0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:
0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂
As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:
0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]
pOH = -log [OH⁻]
pOH = 0.28
And pH = 14 - pOH:
pH = 13.72
Thus, after the addition the pH change from 7.65 to 13.62:
Raise the pH by several units
a) Provide equation of K of this reaction, use symbol " ^ " for exponents. That means 1000 = 10^3 and 1/100 is 10^(-2). b) How many moles of compound F will be produced if only 2 moles of compound C is available? describe or show your work. 3 A + 5 B +4 C 5 D +7 E + F
Answer and Explanation:
a. The equation of K of this reaction is shown below:-
3 A + 5 B + 4 C↔5 D + 7 E + F
[tex]K = \frac{(D)^5 (E)^7 (F)}{(A)^3 (B)^5 (C)^4}[/tex]
b. The moles of compound F is shown below:-
3 A + 5 B + 4 C↔5 D + 7 E + F
2 moles
Now, the mole of produced is
[tex]= \frac{1}{4} \times \ moles\ of\ c[/tex]
Now, we will the value of c by using the above equation
[tex]= \frac{1}{4} \times 2[/tex]
After solving the above equation we will get
0.5 moles
Predict the order of acid strengths in the following series of cationic species: CH3CH2NH3 +, CH3CH=NH2
Answer:
CH3CH=NH2+>CH3CH2NH3 +
Explanation:
There are certain structural features that determine the stability of cationic species. These features that lead to the stability and higher acid strength of cations are those features that stabilize the cation.
CH3CH=NH2+ is more acidic than CH3CH2NH3 + owing to the fact that CH3CH=NH2+ contains a double bond in close proximity with the hydrogen that can be lost as a proton. Electron withdrawal by the double bond (greater s character) makes it easier for this hydrogen to be lost as a proton compared to CH3CH2NH3 +.
1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:
Answer:
1. the siren has a lower pitch to Jane
2. the siren has a higher pitch to John
3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter
Explanation:
The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.
This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.
Describe how you would prepare 500ml of 40% (w/v) aqueous iodine solution.
[Atomic mass of iodine =127g/mol].
Answer:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Explanation:
Hello,
In this case, since 500 mL of a 40% (w/v) aqueous solution iodine is required, we can compute the required mass of iodine by defining the given mass-volume percent:
[tex]\% w/v=\frac{m_{iodine}}{m_{solution}} *100%=\frac{m_{iodine}}{m_{water}+m_{iodine}} *100%[/tex]
In such a way, we need to find mass of iodine, which is computed as:
[tex]m_{iodine}=\frac{\%w/v*m_{water}}{100w/v-\%} \\\\m_{iodine}=\frac{40*500}{100-40}\\ \\m_{iodine}=333.3g\\[/tex]
Thereby, the procedure will be:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Best regards.
Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.
Answer:
A divergent plate boundary
Explanation:
At a divergent boundary, the plates pull away from each other and generate new crust.
Answer:
Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.
Explanation:
PLATO ANSWER
If 75.4 J of energy is absorbed by 0.25 mol of CCl4 at constant pressure, what is the change in temperature? The specific heat of CCl4 is 0.861 J/g·°C.
Answer:
ΔT = 2.28°C
Explanation:
Heat, H = 75.4J
Number of moles = 0.25 mol
Specific heat capacity, c = 0.861 J/g·°C
Change in temperature, ΔT = ?
These quantities are related by the following equation;
H = mc ΔT
Mass, m = Number of moles * Molar mass
m = 0.25mol * 153.82 g/mol
m = 38.455g
S back to the equation;
H = mc ΔT
Substituting the values;
75.4 = 38.455 * 0.861 * ΔT
ΔT = 75.4 / 33.11
ΔT = 2.28°C
The change in temperature is 2.28 °C
First, we will determine the mass of CCl₄ absorbed
From the given information,
Number of moles of CCl₄ absorbed = 0.25 mol
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CCl₄ = 153.82 g/mol
∴ Mass of CCl₄ absorbed = 0.25 × 153.82
Mass of CCl₄ absorbed = 38.455 g
Now, using the formula
Q = mcΔT
Where Q is the quantity of heat
m is the mass
c is the specific heat of substance
and ΔT is the change in temperature
From the given information
Q = 75.4 J
c = 0.861 J/g.°C
Putting the parameters into the formula, we get
75.4 = 38.455 × 0.861 ×ΔT
75.4 = 33.109755 × ΔT
∴ ΔT = 75.4 ÷ 33.109755
ΔT = 2.28 °C
Hence, the change in temperature is 2.28 °C
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How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years
Answer:
68%
Explanation:
Since we need a percentage we can use any number we want for our initial value.
5(1/2)^900/1620 = 3.40
(3.40 / 5)*100 = 68%
To make sure lets use a different initial amount
1(1/2)^900/1620 = 0.68
(0.68/1) * 100 = 68%
To solve this question, we'll assume the initial amount of radium-226 to be 1.
Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:
Step 1Determination of the number of half-lives that has elapsed.
Half-life (t½) = 1620 years
Time (t) = 900 years
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]
Step 2:Determination of the amount remaining
Initial amount (N₀) = 1
Number of half-lives (n) = 5/9
Amount remaining (N) =?[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]
N = 0.68Step 3Determination of the percentage remaining.
Initial amount (N₀) = 1
Amount remaining (N) = 0.68
Percentage remaining =?Percentage remaining = N/N₀ × 100
Percentage remaining = 0.68/1 × 100
Percentage remaining = 68%Therefore, the percentage amount of radium-226 that remains after 900 years is 68%
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What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
For each bond, show the direction of polarity by selecting the correct partial charges. _________ Si-P _________ _________ Si-Cl _________ _________ Cl-P _________ The most polar bond is _______
Answer:
Siδ⁺ -- Pδ⁻⁻
Clδ⁻⁻ -- Pδ⁺
Siδ⁺ -- Clδ⁻⁻
Of the mentioned bonds the most polar bond is Si -- Cl
The polarity of the bond primarily relies upon the electronegativity difference between the two atoms that forms the bond. Therefore, if the electronegativity difference between the two atoms that forms the bond is more the bond will be more polar, and if it is less then the bond will be less polar. The electronegativity of the atoms mentioned is Si = 1.8 , P = 2.1 and Cl = 3.00.
Therefore, the Si - Cl atoms exhibit more electronegativity difference, thus, the Si - Cl bond will be the most polar bond.
How many atoms are in 65.0g of zinc?
from
1moles=iatom
Mole=mass÷avogardos
Where
Avogadro's= 6.02×10²³
So moles = 65.0÷6.02×10²³
Atoms of zinc = 391.6 ×10²³
The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].
Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.
Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately [tex]6.022 \times10^{23}[/tex]atoms per mole.
First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol
Next, multiply the number of moles by Avogadro's number to find the number of atoms.
Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]
Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol
Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.
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A compound is found to contain 29.68 % sulfur and 70.32 % fluorine by mass. What is the empirical formula for this compound?
Answer:
[tex]SF_4[/tex]
Explanation:
The first thing would be to calculate the number of moles of each element in the compound.
No of moles of sulfur (S) = mass/molar mass
= 29.68/32.065 = 0.9256
No of moles of fluorine (F) = mass/molar mass
= 70.32/18.998 = 3.7014
Then, let us find the atomic ratio of each of the element in the compound by dividing by the no of moles by the smallest no of mole:
S : F
[tex]\frac{0.9256}{0.9256}[/tex] = 1 : [tex]\frac{3.7014}{0.9256}[/tex] = 4
Therefore, the empirical formula for the compound is [tex]SF_4[/tex]
An electrolysis cell has two electrodes. Which statement is correct? A. Reduction takes place at the anode, which is positively charged. B. Reduction takes place at the cathode, which is positively charged. C. Reduction takes place at the dynode, which is uncharged. D. Reduction takes place at the cathode, which is negatively charged. E. Reduction takes place at the anode, which is negatively charged.
Answer:
D. Reduction takes place at the cathode, which is negatively charged.
Explanation:
In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.
At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.
Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond.
Answer:
See explanation
Explanation:
The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.
The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.
Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.
The structure that should be drawn is shown below.
The reaction of chlorine:It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.
Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.
What is the concentration of A after 50.7 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 M⁻¹min⁻¹)
Answer:
0.101 M
Explanation:
Step 1: Given data
Initial concentration of A ([A]₀): 0.250 MFinal concentration of A ([A]): ?Time (t): 50.7 minRate constant (k): 0.117 M⁻¹.min⁻¹Step 2: Calculate [A]
For a second-order reaction, we can calculate [A] using the following expression.
1/[A] = 1/[A]₀ + k × t
1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min
[A] = 0.101 M
What are periodic trends if ionic radii
Answer:
Explan ionization energy, atomic radius, and electron affinityation:
This question most likely has answer choices. The possible answer choices are as followed:
Ionic radii tend to increase down a group.Ionic radii tend to decrease across a period.Anionic radii tend to increase across a period.Cationic radii tend to decrease across a period.Ionic radii increase when switching from cations to anions in a period.The answers are Ionic radii tend to increase down a group, Cationic radii tend to decrease across a period, and Ionic radii increase when switching from cations to anions in a period (1st, 4th, and 5th options).
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.
Answer:
[tex]m_{NaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783gH_2[/tex]
Explanation:
Hello,
In this case, the reaction between sodium and ammonia is:
[tex]2Na+2NH_3\rightarrow 2NaNH_2+H_2[/tex]
Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:
[tex]n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa[/tex]
And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):
[tex]n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa[/tex]
In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:
[tex]m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2[/tex]
Best regards.
The following reactions all have K < 1. 1) a. C6H5COO- (aq) + C6H5OH (aq) → C6H5COOH (aq) + C6H5O- (aq) b. F- (aq) + C6H5OH (aq) → C6H5O- (aq) + HF (aq) c. C6H5COOH (aq) + F- (aq) → HF (aq) + C6H5COO- (aq) Arrange the substances based on their relative acid strength.
Answer:
the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
Explanation:
Given that :
a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]
b. [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]
c. [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]
Acid strength is the ability of an acid to dissociate into a proton and an anion. Take for instance.
HA ↔ H⁺ + A⁻
The acid strength of the following compounds above are:
[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰
[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴
[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵
As the acid dissociation constant increases the relative acid strength also increases.
From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid
Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
Why do prices increase when demand for a product is high? Companies know they can make more money by selling fewer products at higher prices. Companies know that people will be willing to spend more to get an in-demand product. Companies take advantage of the demand to make people spend more money on excess products. Companies know they can stop production and still make money on sales.
Answer:
Companies know that people will be willing to spend more to get an in-demand product.
Explanation:
When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.
Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.
Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.
Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.
Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.
Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.
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Consider the reaction for the dissolution of solid magnesium hydroxide.
Mg(OH)2(s)g2 (a) +2OH (ag)
If the concentration of hydroxide ion in a saturated solution of magnesium hydroxide is 2.24 x 104 M.
what is the molar solubility of magnesium hydroxide? Report your answer in scientific notation with three significant figures.
Answer:
Molar solubility is 1.12x10⁻⁴M
Explanation:
The dissolution of magnesium hydroxide is:
Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻
The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:
Mg(OH)₂(s) ⇄ X + 2X
Where X is solubility.
If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:
2X = 2.24x10⁻⁴M
X = 2.24x10⁻⁴M/2
X =1.12x10⁻⁴M
Molar solubility is 1.12x10⁻⁴M
The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?
Answer:
2.8x10⁻¹⁵ M.
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]
In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:
[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]
Thus, solving for [tex]x[/tex] we have:
[tex]x=2.8x10^{-15}M[/tex]
Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.
Regards.
Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(NO3)2. Include physical states.molecular equation:Na_{2}SO_{4}(aq) + Ba(NO_{3})_{2}(aq) ->Na2SO4(aq)+Ba(NO3)2(aq)⟶Enter the balanced net ionic equation for this reaction. Include physical states.net ionic equation:
Answer:
1. The balanced molecular equation is given below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
2. The net ionic equation is given below:
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
Explanation:
1. The balanced molecular equation
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
The above equation can be balance as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
Now, the equation is balanced.
2. The bal net ionic equation.
This can be obtained as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —>
In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:
Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)
Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)
Na2SO4(aq) + Ba(NO3)2(aq) —>
2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)
Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
PLEASE HELP!!
You are performing an experiment that involves the electrolysis of gold (I) bromide, also know as AuBr. You know that gold is less reactive than hydrogen. Which of the following would be the product of the reaction?
A. Hydrogen gas
B. Gold bromide
C. Oxygen gas
D. Pure gold
Answer:
D. Pure gold
Explanation:
Hello,
In this case, since gold, as a heavy metal, is said to be less reactive than hydrogen, when it undergoes electrolysis process when forming a salt, due to the action of the electric current, we can appreciate the formation of a layer of gold on the surface of the cathode via a reduction half-reaction from gold (I) to metallic gold:
[tex]Au^++1e^-\rightarrow Au^0[/tex]
Thereby, D. Pure gold is formed as the product of the reaction.
In contrast, more reactive metals than hydrogen such as sodium or potassium, will remain in solution so the hydrogen converted to hydrogen gas.
Best regards-
Most of the costs associated with using renewable resources are due to
а. overuse of resources
b. atmospheric pollution
C.lack of availability
d.global warming
Answer:
The answer is a.
Explanation:
Most of the costs associated with using renewable resources are due to overuse of the resources.
Most of the costs associated with using renewable resources are due to overuse of resources.
What are renewable resources?Renewable resources are those resources which will be generated naturally and continously from the nature and these are also inexhaustible means non ended.
As from the definition it is clear that we can reuse or will use again and again these types of resources, that's why cost associated with these renewable resources are high.
Atmospheric pollution and global warming causes hazardous effect on the environment, so it will not be the reason with the associated cost.Lack of availability makes its important, not costly and in our daily life we used many kinds of renewable resorces so it is not possible to use costly resources daily.Hence, overuse of resouces is one of the reason.
To know more about renewable resorces, visit the below link:
https://brainly.com/question/79953
The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.
Answer:
The answer is
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]Explanation:
The Ka of an acid when given the pH and concentration can be found by
[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]where
c is the concentration of the acid
From the question
pH = 5.82
c = 0.010 M
Substitute the values into the above formula and solve for Ka
We have
[tex]5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010) [/tex][tex] - \frac{1}{2} log(Ka) = 5.82 + 1[/tex][tex] - \frac{1}{2} log(Ka) = 6.82[/tex]Multiply through by - 2
[tex] log(Ka) = - 13.64[/tex]Find antilog of both sides
We have the final answer as
[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]Hope this helps you
Which of the following is a covalent bond? A NaCl B K2O C H2O D MgO
Answer:
[tex]H_2O[/tex]
Explanation: Research has proven that ;
Water is a Polar Covalent Molecule
It consists of 2 Hydrogen molecules bonded to one Oxygen molecule and the two hydrogen atoms are not evenly distributed around the oxygen atom.
Write a balanced nuclear equation for the following: The nuclide thorium-234 undergoes beta decay to form protactinium-234 .
Answer:
²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e
Explanation:
thorium-234 = ²³⁴₉₀Th
beta decay = ⁰₋₁e
protactinium-234 = ²³⁴₉₁Pa
The balanced nuclear equation is given as;
²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e