Answer:
0.0347 moles of hydronium ions
Explanation:
The equation of the neutralization reaction between hydroxide and hydronium ions is given below:
H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)
From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.
The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:
Number of moles = mass / molar mass
Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol
Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol
Mass of magnesium hydroxide = 450 g = 0.45 g
Mass of aluminium hydroxide = 500 mg = 0.5 g
Moles of magnesium hydroxide = (0.45/58) moles
Moles of aluminium hydroxide = (0.5/78) moles
Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:
Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)
Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)
Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles
Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles
Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions
Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
________________________________________________________
NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.
How we calculate the [OH⁻]?We can calculate the concentration of hydroxide ions as follow:
[OH⁻][H⁺] = 10⁻¹⁴
Given chemical reaction with ICE table shown as below:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq)
Initial: 0.115 0 0
Change: -x +x +x
Equilibrium: 0.115-x +x +x
Given that, Ka = 4.3 x 10⁻⁷
Equilibrium constant for this reaction is written as:
Ka = [H⁺][HCO₃⁻]/[H₂CO₃]
4.3 x 10⁻⁷ = x² / 0.115
x = 2.32 x 10⁻⁴M = [H⁺]
Now we calculate the concentration of hydroxide ion as:
[OH⁻][H⁺] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M
Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.
To know more about pH & pOH, visit the below ink:
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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.
Answer:
[tex]T_2= 36.7 \textdegree C[/tex]
Explanation:
Mass of Water [tex]m_w=6.90kg[/tex]
Temperature [tex]T=34.7 degrees[/tex]
Heat Flow [tex]H=57.1kJ[/tex]
Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]
Generally the equation for Final Temperature is mathematically given by
[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]
[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]
Therefore
[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]
[tex]T_2= 36.7 \textdegree C[/tex]
For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-1 s -1 . If the initial concentration was 2.2 M, calculate the concentration after 100 seconds. Show your work.
Answer:
[tex]C_t=0.165M[/tex]
Explanation:
From the question we are told that:
Slope [tex]K=0.056 M-1 s -1[/tex]
initial Concentration [tex]C_1=2.2M[/tex]
Time [tex]t=100[/tex]
Generally the equation for Raw law is mathematically given by
[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]
[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]
[tex]C_t=0.165M[/tex]
The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.
What is the specific rate constant?The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.
Given,
Slope (k)= 0.056 per M per s
Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M
Time (t) = 100 seconds
The concentration of the reaction after 100 seconds can be given by,
[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]
Substitute values in the above equation:
[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]
Therefore, after 100 seconds the concentration is 0.165 M.
Learn more about the order of reaction here:
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What is the equilibrium expression for the reaction below?
Caco (s)
Cao(s) + CO (g)
A.
B.
[Cacoz]
[Cao]
[Caco.]
[Cao]+[CO,
[Cao][COL]
[Caco:]
C.
D. [co]
Answer:
D. [CO₂]
Explanation:
Let's consider the following equation at equilibrium.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [CO₂]
all question are compulsory
Answer:
is this question or you just asking I can't understand.
The forensic technician at a crime scene has just prepared a luminol stock solution by adding 17.0 \({\rm g}\) of luminol into a total volume of 75.0 \(\rm mL\) of \(\rm H_2O\). What is the molarity of the stock solution of luminol
Answer:
1.28 M
Explanation:
Step 1: Given data
Mass of luminol (solute): 17.0 g
Volume of water: 75.0 mL (this is also the volume of solution)
Step 2: Calculate the moles corresponding to 17.0 g of luminol
The molar mass of luminol is 177.16 g/mol.
17.0 g × 1 mol/177.16 g = 0.0960 mol
Step 3: Calculate the molarity of the solution
We will use the definition of molarity
M = moles of solute / liters of solution
M = 0.0960 mol / 0.0750 L = 1.28 M
A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?
Answer:
6.684g
Explanation:
Here, we can use the mole ratio of the gases to calculate.
We know that the mole ratio of the gases equate to their number of moles.
Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol
Thus, the number of moles produced is 5.98/32 = 0.186875
Where do we move from here?
We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.
449/851 = 0.186875/n
n =(0.186875 * 851)/449
n = 0.3542
Now we do the same for argon to get the number of moles of argon.
Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.
P = 851 - 449 = 402 mmHg
We now use the mole ratio relation.
402/851 = n/0.3542
n = (402 * 0.3542) / 851
n = 0.1673
Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.
The atomic mass of argon is 39.948 amu
The mass is thus 39.948 * 0.1673 = 6.684g
write down the different uses of water that you know about
Answer:
The various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water is used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Answer:
The various use of water are;
I) Cooking.
ii) Drinking
III) Bathing
iv) Generating hydro- electricity
v) Construction work etc
Write a chemical equation for LiOH(aq) showing how it is an acid or a base according to the Arrhenius definition.
Answer:
LiOH(aq) → Li⁺(aq) + OH⁻(aq).
Consider the following titration for these three questions:
1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.
a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?
Answer:
a. 1 mole of acid is equal to one equivalent.
b. 1.00 moles of HCl are found.
c. 1L of 2.00M NaOH is needed to reach the equivalence point
Explanation:
HCl reacts with NaOH as follows:
HCl + NaOH → NaCl + H2O
Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1
a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent
b. The initial moles of HCl are:
1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl
At the halfway point, the moles of HCl are the half, that is:
1.00 moles of HCl are found
c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:
2.00moles NaOH * (1L / 2.00mol) =
1L of 2.00M NaOH is needed to reach the equivalence point
Which type of organic compound is shown below?
A. Carboxylic acid
B. Ester
C. Amine
D. Alcohol
Answer:
I think its A maybe am not sure
work 10
HOMEWORK ASSIGNMENTS Content
Detai
Question 1
6.25 Points
р А.
71
Calculate AE of a gas for a process in which the gas evolves 24 J of heat and has 9 of work done on it.
A
✓
A -33)
B +33)
Gradin
-220)
D) +15)
E -15)
Question 2
6.25 Points
Answer
A
Explanation:
due to high specific heat capacity it loses heat and has low temperature
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is started with 2.6 grams of NH3 (g) added into a reaction vessel with CO2 (g).
Write the balanced equation for this reaction, being sure to include physical states. Based on the balanced equation above, calculate the following:
a. the theoretical yield of urea in grams that can be made from the NH3
b. the actual amount of urea made if the percent yield for this reaction is 34%.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
You are titrating 24.3 mL of 2.00 M HCl with 1.87 M NaOH. How much NaOH do you expect to have added when you reach the equivalence point?
26.0 mL
15.4 mL
13.4 mL
Answer:
26mL
Explanation:
NaOH+HCl= NaCl+H2O
nHCl=0.0243*2=0.0486
nNaOh=nHCl
VNaOH=0.0486/1.87=0.026l=26ml
Answer:
26.0 mL
Explanation:
For the following reaction, 5.29 grams of water are mixed with excess diphosphorus pentoxide. The reaction yields 13.3 grams of phosphoric acid . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq). What is the theoretical yield of phosphoric acid?
Answer:
19.2 g
Explanation:
Step 1: Write the balanced equation
P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄
Step 2: Calculate the moles corresponding to 5.29 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
5.29 g × 1 mol/18.02 g = 0.294 mol
Step 3: Calculate the theoretical yield of phosphoric acid, in moles
The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol
Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
0.196 mol × 97.99 g/mol = 19.2 g
# of protons
# of neutrons
# of electrons
Atomic Number
Mass Number
18
17
35
17
37
6
8
6
6
15
Answer:
35
Explanation:
is the answer for your question
The correct geometry around oxygen in CH3OCH3 is
(a). linear. (b). bent. C). tetrahedral/(a). trigonal planar
Explanation:
the force of the lone pairs from the bottom would cancel out the force of the lone pairs from the top. Thus, the molecule will be linear.
Critique this statement: Electrons can exist in any position
outside of the nucleus.
Answer:
However, there has to be 2 electrons on the first shell, and 8 on the others.
Explanation:
Hope this helps :)
Select the correct relationship among the concentrations of species present in a 1.0 M aqueous solution of the weak acid represented by HA. A. [H2O] > [HA] > [A-] > [H3O ] > [OH-] B. [H2O] > [A-] ~ [H3O ] > [HA] > [OH-] C. [HA] > [H2O] > [A-] > [H3O ] > [OH-] D. [H2O] > [HA] > [A-] ~ [H3O ] > [OH-] E. [HA] > [H2O] > [A-] ~ [H3O ] > [OH-]
Answer:
D
Explanation:
We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.
Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.
Hence, for this solution, the concentration of ions in solution follows the order;
[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]
A student was asked to determine the percent mass of sodium nitrate in a mixture of sodium nitrate (NaNO3) and calcium carbonate (CaCO3). The mass of the mixture used was 3.2 g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the solution by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighted 0.45 g. The student dried the insoluble residue of CaCO3 and found that it weighted 2.23 g. Calculate the percent mass of NaNO3 in the mixture and round your final answer to the correct number of significant figures.
Answer:
自分の仕事をする translate to english
Explanation:
list two uses of H2SO4
PLEASE HELP I NEED THIS ASAP
Select all that apply.
The spectrum of Star S is compared to a reference hydrogen spectrum. What can be concluded about Star S?
Star S has radial motion.
Star S has transverse motion.
Star S is moving toward Earth.
Star S is moving away from Earth.
Answer:
I say Star S has radial motion
Explanation:
I'm not sure if it right but let me know if you have any other questions
The half life for the decay of carbon-14 is 5.73 times 10^3 years.
Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of wood from an archeological dig is measured to be 77.The activity in a similar-sized sample of fresh wood is measured to be 85.Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
790 years
Explanation:
Given that;
0.693/t1/2 = 2.303/t log [A]o/[A]
So;
t1/2 =half life of carbon-14
t= age of the sample
[A]o= activity of the living sampoke
[A] = activity at time t
0.693/5.73 ×10^3 = 2.303/t log 85/77
1.21 × 10^-4 = 2.303/t log 1.1
1.21 × 10^-4 = 0.0953/t
t= 0.0953/1.21 × 10^-4
t= 790 years (to 2sf)
We know that water is purified before it is supplied to our houses. Then why do we have filters installed in our houses? What do they serve?
Answer:
Water filters remove elements that cause drinking water to have an unpleasant taste and smell, such as lead, chlorine and bacteria. Home water filtration system will improve the overall purity, taste and smell of your drinking water. It also lowers the pH level of the water that you drink.
Kamal was told by his mother to pour water through a thin cloth into another container to further purify the water.
a) What do you think will happen to the mud and sand when pouring the water? b)Do you think the water filtered by Kamal is safe to drink?
a].When sand is added to water it either hangs in the water or forms a layer at the bottom of the container. Sand therefore does not dissolve in water and is insoluble. It is easy to separate sand and water by filtering the mixture.
b]. The water filtered by kamal is not safe to drink .
If you run out of water, or cannot carry enough water with you for your entire trip, you may need to source drinking water from natural water sources.
The Nernst equation at 20oC is:
Eion= 58 millvolts/z. [log10 (ion)out/(ion)in]
Calculate the equilibrium potential for Cl- if the concentration of Cl- outside of the cell is 100 and the concentration inside of the cell is 10 mmol/liter.
a. 58 millivolts
b. +58 millivolts
c. -116 millivolts
d. 0
Answer:
a. -58 millivolts
Explanation:
The given Nernst equation is:
[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]
The equilibrium potential given by the Nernst equation can be determined by using the formula:
[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]
where:
gas constant(R) = 8.314 J/K/mol
Temperature (T) = (20+273)K
= 298K
Faraday constant F = 96485 C/mol
Number of electron on Cl = -1
[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]
[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]
[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]
[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion: Fe3+(aq) + 3 C2O42-(aq) ⇌ [Fe(C2O4)3]3-(aq). If 0.0800 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium?
Answer:
hello? are you still here? reply if you are
3. Suppose you wanted to design an experiment to test the composition of a mixture that includes sodium phenoxide (NaC6H5O). You know that this solid mixture contains both the NaC6H5O and some inert NaCl, but do not know how much of each is present. You decide to test the composition by titrating with 0.100-M HCl. a. If a 1.000-g sample is 25% NaC6H5O by mass, how many mL of 0.100-M HCl would be required to reach the equivalence point of the titration
Answer:
21.5mL of a 0.100M HCl are required
Explanation:
The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.
To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:
Mass NaC6H5O:
1.000g * 25% = 0.250g NaC6H5O
Moles NaC6H5O -116.09g/mol-
0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required
Volume 0.100M HCl:
2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =
21.5mL of a 0.100M HCl are requiredHow many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
