Answer:
A mixture of 3 moles of N2, 5 moles of CO2, and 10moles of Cl2 exert a total pressure of 1120 mmHg. What is the partial pressure of CO2?
Explanation:
According to Dalton's law of partial pressures:
The partial pressure of a gas can be determined by using the formula:
[tex]the partial pressure of a gas = mole fraction of the gas * total pressure[/tex]
Partial pressure of CO2:
[tex]partial pressure of CO2= total pressure * mole fraction of CO2\\\\Mole fraction of CO2=\frac{number of moles of CO2}{total number of moles of all the gases} \\mole fraction of CO2=\frac{5mol}{3mol+5mol+10mol} = 5/18\\Partial pressure of CO2=\frac{5}{18} * 1120mmHg\\ =311.1mmHg[/tex]
Hence, the partial pressure of CO2 is 311.1mmHg.
The relationship between the volume and mass of an element is
Answer:
Density
Explanation:
The mass of an element is the average weight that the isotope of the particular element contains. Its characteristic indicates the amount of substance present in an element.
However, the volume of an element on the other hand is the mole of an element that is contained in a room temperature.
The relation joining both the mass and volume of an element is density.
This is because density showcase the relationship between the mass of an element to space in occupies in terms of volume.
It is given by the formula:
Density = mass/volume
Which of the choices below has more heat being transferred as thermal energy from one place to another?
A. A bowl of ice water
B. A pot of boiling water
Answer:
B
Explanation:
So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.
How many moles of electrons must be transferred to plate out 110 g of manganese (MW ~ 55g/mol) from a solution of permanganate (MnO4-)
Answer:
14 mol e⁻
Explanation:
Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese
8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)
Step 2: Calculate the moles corresponding to 110 g of manganese
The molar mass of Mn is 55 g/mol.
110 g × 1 mol/55 g = 2 mol
Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn
According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.
2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻
A rock was weighed on a balance and found to have a mass of 34.56 grams. It was placed into water that was in a graduated cylinder. Before the rock was dropped into the water the water level was 22.7 mL. The addition of the solid object caused the water level to rise to 28.1 mL. What is the density of the rock?
Answer:
5 and the rest are all set to the same date on your list as the other one to get you a list on for a your special first year week and with a special holiday party holiday
Explanation:
Sorry desperate for points
Which hydrocarbon is saturated?
A)
heptyne
B)
butane
C)
propene
D)
ethyne
163 mL of 2.75 mol/L aluminum sulfate is to have all aluminum ions removed by adding a 3.65 mol/L sodium sulfide solution.
Al2(SO4)3(aq) + 3 Na2S(aq) → Al2S3(s) + 3 Na2SO4(aq)
What is the minimum volume of sodium sulfide that must be added to completely remove all the aluminum ions? What mass of aluminum sulfide will form?
Answer:
[tex]V_{Na_2S}=368mL[/tex]
[tex]m_{Al_2S_3}=67.3gAl_2S_3[/tex]
Explanation:
Hello there!
In this case, according to the given information, it is possible to realize that the only way for us to calculate the required volume of sodium sulfide, is by calculating the moles of this substance consumed 163 mL of 2.75 mol/L aluminum sulfate by using the definition of molar concentration and the 1:3 mole ratio between these two:
[tex]n_{Na_2S}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{3molNa_2S}{1molAl_2(SO_4)_3} =1.34molNa_2S[/tex]
Now, we divide these moles by the molar concentration of sodium sulfide to obtain the required volume:
[tex]V_{Na_2S}=\frac{1.34molNa_2S}{3.65mol/L} =0.368L=368mL[/tex]
For the last part, we now use the 1:1 mole ratio of aluminum sulfate to aluminum sulfide and the molar mass of the latter (150.158 g/mol) in order to calculate the required mass:
[tex]m_{Al_2S_3}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{1molAl_2S_3}{1molAl_2(SO_4)_3} *\frac{150.158gAl_2S_3}{1molAl_2S_3} \\\\m_{Al_2S_3}=67.3gAl_2S_3[/tex]
Regards!
Hydrogen gas was cooled from 150 K to 50 K. Its new volume (V2) is 75 mL. What was its original volume (V1)?
Explanation:
57.3ml
we use Charles's law
to solve the question
Answer:
[tex]\boxed {\boxed {\sf 225 \ mL}}[/tex]
Explanation:
The temperature and volume of the gas are changing, so we use Charles's Law. This states the temperature of a gas is directly proportional to the volume of a gas. The formula is:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
The original volume is unknown. The new volume is 75 milliliters.
The gas is cooled from 150 Kelvin to 50 Kelvin, so the original temperature is 150 K and the new temperature is 50 K.
We know that:
T₁= 150 K V₂= 75 mL T₂= 50 KSubstitute the values into the formula.
[tex]\frac {V_1}{150 \ K}=\frac{ 75 \ mL}{50 \ K}[/tex]
Since we are solving for the original volume, we must isolate the variable V₁.
It is being divided by 150 K. The inverse of division is multiplication, so we multiply both sides by 150 K.
[tex]150 \ K *\frac {V_1}{150 \ K}=\frac{ 75 \ mL}{50 \ K}* 150 \ K[/tex]
[tex]V_1=\frac{ 75 \ mL}{50 \ K}* 150 \ K[/tex]
The units of Kelvin (K) cancel.
[tex]V_1= \frac{ 75 \ mL}{50 }* 150[/tex]
[tex]V_1=1.5 * 150 \ mL[/tex]
[tex]V_1= 225 \ mL[/tex]
The original volume is 225 milliliters.
In what form is energy that is NOT used for life processes released from living things?
1. Thermal energy
2. Chemical energy
3. Light energy
4. Sound energy
Answer:
Correct answer would be Option 2, Chemical Energy
Hope this helps!
What is the molarity if 6664 grams of boron iodide is dissolved in 5.25 liters of water?
Answer:
3.24 mol/L
Explanation:
Given that:
mass of Boron triiodide = 6664 grams
molar mass of BI_3 = 391.52 g/mol
Recall that:
number of moles = mass/molar mass
∴
number of moles = 6664 g /391.52 g/mol
number of moles = 17.02 mol
Also;
Molarity = moles for solute/liter for solution
= 17.02 mol/5.25 L
= 3.24 mol/L
Order the following chemicals from lowest to highest melting point.
Tungsten
Copper (II) chloride
• Caffeine
• Lead
• Sodium chloride
• Silver
Answer
• Silver
• copper
How to change τhe color of silver mercury to red???
Answer: This is the hue of mercury if you're looking for it. Mercury reacts with sulphur to generate a crimson hue. In the past, they always used a thermometer.
Explanation: However, due of its toxicity, individuals have switched to using alcohol in glass thermometers, which work similarly to mercury thermometers.
hope this helped best of luck mate! :) if this helped make sure to mark me Brainliest!
What's 2+7 hehdhenbeidbnejdbdbdbdi
Answer:
Answer
9......
Explanation:
Explaination:
True or false? An organism may play more than one role in a food web
Answer:
True
Explanation:
Your welcome! :) Good luck!
To run a spectrophotometry experiment, begin by _______ the spectrophotometer and preparing the samples. Be sure to select the correct ________, then run a measurement on the _______ solution. Follow up by running measurements on _______ solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.
1.
a) cleaning
b) warming up
c) unplugging
2.
a) wavelength
b) transmittance
c) absorbance
3.
a) blank
b) aqueous
c) sample
4.
a) blank
b) aqueous
c) sample
Answer:
b) warming up a) wavelength a) blank c) sample
Explanation:
To run a spectrophotometry experiment, begin by warming up the spectrophotometer and preparing the samples. It is important that the equipment is warmed up for at least 30 minutes before starting the measurements.
Be sure to select the correct wavelength, then run a measurement on the blank solution. The selected wavelength depends on the analyte of interest. The black solution contains the same matrix but it doesn´t contain the analyte.
Follow up by running measurements on sample solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples. The samples are those of unknown concentration that we want to determine.
To run a spectrophotometry experiment, begin by cleaning the spectrophotometer and preparing the samples. Be sure to select the correct wavelength, then run a measurement on the sample solution. Follow up by running measurements on aqueous solutions. Once data is collected, turn off the instrument, clean the area, and discard the samples.
Spectrophotometry is a technique used to measure the interaction of light with matter, specifically the absorption, transmission, or reflection of light by a sample. It involves the use of a spectrophotometer, an instrument that measures the intensity of light as a function of its wavelength or frequency.
In spectrophotometry, a sample is exposed to light of a specific wavelength or a range of wavelengths. The sample may absorb certain wavelengths of light, which can be detected and measured by the spectrophotometer. The amount of light absorbed is related to the concentration of the analyte in the sample, allowing for quantitative analysis.
Learn more about Spectrophotometry, here:
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Please help me complete this, it’s my last chance
Protons: charge: +1 // mass: 1 // location: nucleus
Neutrons: charge: 0// mass: 1 // location: nucleus
Electrons: charge: -1// mass: 0// location: orbitals
8. What does the term 'sustainable mean? *
Answer:
something that can be maintained over a period of time
Answer:
a balance between meeting today's needs.......
Explanation:
Which substance has Delta.Hf defined as 0 kJ/mol? H2O (s) Ne (l) F2 (g) CO2 (g)
Answer:
F2 (g)
Explanation:
Edg 2021
Answer:
F2 g
Explanation:
The temperature of a sample of silver increased by 23.8 °C
when 261 J of heat was applied.
What is the mass of the sample?
Answer:
46.67 g
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 23.8 °C
Heat (Q) = 261 J
Specific heat capacity (C) of silver = 0.235 J/gºC
Mass of silver (M) =?
The mass of the sample of silver can be obtained as follow:
Q = MCΔT
261 = M × 0.235 × 23.8
261 = M × 5.593
Divide both side by 5.593
M = 261 / 5.593
M = 46.67 g
Thus, the mass of the sample of silver is 46.67 g
I want to know which ones are molecular equation, complete ionic equation and net ionic equation
Answer:
The molecular equations are:
1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)
2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)
The complete ionic equations are:
1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)
2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)
The net ionic equations are:
1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)
2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)
Explanation:
A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.
An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.
In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.
From the given attachment;
The molecular equations are:
1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)
2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)
The complete ionic equations are:
1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)
2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)
The net ionic equations are:
1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)
2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)
Silver sulfate is precipitated from solution according to the following reaction 2 AgNO3 (aq) Na2SO4 (aq) --> Ag2SO4 (s) 2 NaNO3 (aq) If 3.80 moles of AgNO3 and 3.05 moles of Na2SO4 are present initially, how many moles of excess reactant will remain after precipitation is complete
Answer:
1.15 moles of excess reactant will remain after precipitation is complete.
Explanation:
The balanced reaction is:
2 AgNO₃ (aq) + Na₂SO₄ (aq) → Ag₂SO₄ (s) + 2 NaNO₃ (aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
AgNO₃: 2 molesNa₂SO₄: 1 moleAg₂SO₄: 1 moleNaNO₃: 2 molesThen you can apply the following rule of three:: if by stoichiometry 2 moles of AgNO₃ reacts with 1 mole of Na₂SO₄, 3.80 moles of AgNO₃ reacts with how much moles of Na₂SO₄?
[tex]amount of moles of Na_{2}SO_{4} =\frac{1mole of Na_{2}SO_{4} * 3.80 moles of AgNO_{3} }{2 mols of AgNO_{3} }[/tex]
amount of moles of Na₂SO₄= 1.9 moles
But 3.05 moles of Na₂SO₄ are available. Since you have more moles than you need to react with 3.80 moles of AgNO₃, Na₂SO₄ will be the excess reagent.
To calculate the amount of excess reagent that will remain, you must make the difference between the amount you initially have and the amount that reacts:
3.05 moles - 1.9 moles= 1.15 moles
1.15 moles of excess reactant will remain after precipitation is complete.
Use the graph to calculate the instantaneous rate of formation of HBr at 50 s
Express your answer using one significant figure.
Answer: The instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]
Explanation:
From the graph,
Initial rate of the [tex]Br_2[/tex] = 1.0 M
Time when the concentration of [tex]Br_2[/tex] is 0.5 M (half the concentration ) = 60 sec
For first order reaction:
Calculating rate constant for first order reaction using half life:
[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)
[tex]t_{1/2}[/tex] = half life period = 60 s
k = rate constant = ?
Putting values in equation 1:
[tex]k=\frac{0.693}{60s}\\\\k=0.01155s^{-1}[/tex]
For the given chemical reaction:
[tex]H_2(g)+Br_2(g)\rightarrow 2HBr(g)[/tex]
Rate of the reaction = [tex]-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]
Negative sign represents the disappearance of the reactants
From the above expression:
[tex]k[Br_2]=-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]
At 50 seconds, [tex][Br_2]=0.6 M[/tex]
Plugging values in above expression, we get:
[tex]\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=0.01155\times 0.6\\\\\frac{\Delta [HBr]}{\Delta t}=2\times 0.01155\times 0.6=0.01386=1.4\times 10^{-2}M/s[/tex]
Hence, the instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]
2. Energy cycles through ecosystems because it returns to the sun.
True or false
Please help I will give 50 points
Answer:
false
Explanation:
energy does not return to the sun, it returns to the plants or producers. if energy were to return to the sun, it would have to travel though space.
rank in order the strongest to the weakest acid cbr3cooh, ch3cooh and ccl3cooh
Answer: Rank in order the strongest to the weakest acid is [tex]CCl_{3}COOH[/tex] > [tex]CBr_{3}COOH[/tex] > [tex]CH_{3}COOH[/tex].
Explanation:
More readily a substance is able to donate a hydrogen ion more will be its acidic strength. Hence, stronger will be the acid.
More is the electronegativity of atoms attached to the acid more easily it will donate a proton. Hence, more will be its acidic strength.
Chlorine is more electronegative in nature as compared to bromine. So,
[tex]CCl_{3}COOH[/tex] is more acidic than [tex]CBr_{3}COOH[/tex].
Since there is no electronegative group attached to [tex]CH_{3}COOH[/tex] so it is least acidic than [tex]CCl_{3}COOH[/tex] and [tex]CBr_{3}COOH[/tex].
Thus, we can conclude that rank in order the strongest to the weakest acid is [tex]CCl_{3}COOH[/tex] > [tex]CBr_{3}COOH[/tex] > [tex]CH_{3}COOH[/tex].
You are given a 250 ml sample of HCI and asked to find out what its concentration is. You have a 0 118 M Ca(OH)2 solution and it takes 13.7 ml to
neutralize the acid sample. What is the concentration of the HCI? Show explanation Please help this is urgent
Answer:
The concentration of the acid, HCl is 0.013 M
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O
From the balanced equation above,
The mole ratio of the acid, HCl (nₐ) = 2
The mole ratio of base, Ca(OH)₂ (n₆) = 1
Finally, we shall determine the molarity of the HCl. This can be obtained as follow:
Volume of acid, HCl (Vₐ) = 250 mL
Molarity of base, Ca(OH)₂ (M₆) = 0.118 M
Volume of base, Ca(OH)₂ (V₆) = 13.7 mL
Molarity of acid, HCl (Mₐ) =?
MₐVₐ / M₆V₆ = nₐ/n₆
Mₐ × 250 / 0.118 × 13.7 = 2/1
Mₐ × 250 / 1.6166 = 2
Cross multiply
Mₐ × 250 = 1.6166 × 2
Mₐ × 250 = 3.2332
Divide both side by side 250
Mₐ = 3.2332 / 250
Mₐ = 0.013 M
Thus, the concentration of the acid, HCl is 0.013 M
so de fanga gu se si gage humang we sapa shata phoni we heo sang pai
Answer:
Which language is this???
A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. The acid required 27.4 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?
Solution :
The equation is :
[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]
The number of the moles of HA os 0.00285, and the volume is 25 mL.
15 mL of the 0.0950 M NaOH is added.
The total volume of a solution is V = 25 mL + 15 mL = 40 mL
The pH of the solution is 6.50
Calculating the [tex]K_a[/tex] of HA
[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
Let s calculate the concentration of HA and NaOH
[tex]$[HA] = \frac{^nH_A}{V}$[/tex]
[tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]
= 0.07125 M
[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]
[tex]$=\frac{0.001425 mol}{0.04L}$[/tex]
= 0.0356 M
[tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]
Initial conc. (M) 0.07125 M 0.0356 M 0 M
Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M
Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M
Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M
0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]
Now for [tex][H^+][/tex]
[tex]$[H^+] = 10^{-pH}$[/tex]
[tex]$=10^{-6.5}$[/tex]
[tex]$=3.16 \times 10^{-7}$[/tex]
Calculating the value of [tex]K_a[/tex],
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
[tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]
[tex]$=3.16\times 10^{-7}$[/tex]
Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].
Of the below gases, which would deviate most from ideal gas behavior? CO O2 NH3 SF4
Answer:
For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality.
Explanation:
Calculate the number of milliliters of 0.587 M NaOH required to precipitate all of the Fe3 ions in 197 mL of 0.654 M FeCl3 solution as Fe(OH)3. The equation for the reaction is: FeCl3(aq) 3NaOH(aq) Fe(OH)3(s) 3NaCl(aq)
Answer: The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].
Explanation:
The reaction equation is as follows.
[tex]FeCl_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaCl(aq)[/tex]
Therefore, moles of [tex]Fe(OH)_{3}[/tex] are calculated as follows.
Moles = Molarity of [tex]Fe(OH)_{3}[/tex] [tex]\times[/tex] Volume (in L)
= 0.654 M [tex]\times[/tex] 0.197 L
= 0.128 mol
Now, according to the given balanced equation 1 mole of [tex]FeCl_{3}(aq)[/tex] reacts with 3 moles of NaOH(aq). Hence, moles of [tex]Fe(OH)_{3}[/tex] reacted are calculated as follows.
3 [tex]\times[/tex] 0.128 mol = 0.384 moles of NaOH
As moles of NaOH present are as follows.
Moles of NaOH = Molarity of NaOH [tex]\times[/tex] Volume (in L)
0.384 mol = 0.587 M [tex]\times[/tex] Volume (in L)
Volume (in L) = 0.654 L (1 L = 1000 mL) = 654 mL
Thus, we can conclude that the number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].
You deserve 0.72G of potassium chloride (KCI) in 600 ml of water. What is the molarity of the solution ?
Answer:
0.016 M
Explanation:
Molarity refers to the molar concentration of a solution and it can be calculated using the formula below:
Molarity (M) = number of moles (n) ÷ volume (V)
According to this question, the mass of KCl was given to be 0.72 grams and the volume of water as 600 mL.
Using mole = mass/molar mass to convert mass of KCl to moles
Molar mass of KCl = 39 + 35.5 = 74.5g/mol
mole = 0.72g ÷ 74.5g/mol
mole = 0.00966mol
Volume of water = 600mL = 600/1000 = 0.600L
Molarity, M = 0.00966 ÷ 0.600
Molarity of KCl solution = 0.016 M
Rank the compounds below in order of decreasing base strength.
a. HPO4^2-
b. NH3
c. NO2
d. SO4^2-
Answer:
c. NO2
b. NH3
d. SO4 ^2 -
a. HPO4 ^ 2 -
Explanation:
Acid is a compound which ionizes to produce hydrogen ions. The Ph value for acid is below 7. Base is a compound which ionizes to produce hydroxide ions. The Ph value for base is above 7. The Conjugate base accepts a proton or releases a hydrogen ion.