Answer:
0.043 M
Explanation:
The reaction that takes place is:
Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂OFirst we calculate how many HCl moles reacted, using the given concentration and volume required to reach the equivalence point:
0.029 M HCl * 37.3 mL = 1.0817 mmol HCl = 1.0817 mmol H⁺As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.
With that in mind we can calculate the hydroxide ion concentration in the original sample solution, using the calculated number of moles and given volume:
1.0817 mmol OH⁻ / 25.0 mL = 0.043 MThe titration is termed the neutralization reaction with the acid and base. The concentration of hydroxide in the titration is 0.0865 M.
What is a neutralization reaction?The neutralization reaction is given as the reaction in which the acid and base react to form the salt and water, stabilizing the pH of the solution.
The neutralization of acid and base to identify the strength can be given as:
[tex]\rm M_V_1=M_2V_2[/tex]
Substituting the strength and the volume of calcium hydroxide and the HCl with the volume:
[tex]\rm 0.029\;M\;\times\;37.3\;mL=M_2\;\times\;25\;mL\\M_2=\dfrac{ 0.029\;M\;\times\;37.3\;mL}{25\;mL} \\M_2=0.0432\;M[/tex]
The strength of the calcium hydroxide in the reaction is 0.04326 M.
One molar unit of calcium hydroxide results in 2 molar units of hydroxide. The molar unit of hydroxide in 0.04326 M calcium hydroxide is:
[tex]\rm 1\;M\;Ca(OH)_2=2\;M\;OH^-\\\\0.04326\;M\;Ca(OH)_2=0.04326\;\times\;2\;M\;OH^-\\0.04326\;M\;Ca(OH)_2=0.0865\;M\;OH^-[/tex]
The concentration of hydroxide ion in the titration is 0.0865 M.
Learn more about molarity, here:
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Give the name of the molecular compound and the name of the aqueous solution for each of the binary compounds below.
a. HCN
b. HF
c. H2S
Explanation:
A compound containing only two elements is called a binary compound.
(a). The compound HCN contains a hydrogen ion and cyanide ion. Hence, the name of HCN is hydrogen cyanide.
An aqueous solution of hydrogen cyanide is known as hydrogen cyanide solution.
(b). The compound HF contains a hydrogen ion and a fluorine ion. The suffix -ide will be added to its name. So, the name of this compound is hydrogen fluoride.
An aqueous solution of HF is known as hydrogen fluoride.
(c). The compound [tex]H_{2}S[/tex] contains two hydrogen atoms and one sulfur atom. The suffix -ide will be added to its name. So, the name of this compound is hydrogen sulfide.
An aqueous solution of this compound is known as hydrogen sulfide solution.
1.Using the absorbance of the spinach extract and the equation of the trendline, determine the concentration of the extract solution. Report the concentration in moles/L (M).
2. Calculate the number of rams of chlorophyll-a in the 25ml, spinach solution.
3. Calculate the concentration of the chloropyhll-a soultion in spinach (mg chlorophyll-a/g spinach)
Trendline: y=1609x + .0055
Absorbance spinach extract lamda max: .329
Absorbacne spinach extract, 750 nm: .023
Corrected absorbance: .306
Mass of Spinach: .1876g Total Volume of spinach: 25mL
Answer:
Explanation:
From the given information:
We are to make use of the spinach absorbance extract which is the corrected absorbance (y) = 0.306
And also the trendline equation:
y = 1609x + 0.0055
where,
x = absorbance of the spinach extract.
∴
0.306 = 1609x + 0.0055
collecting the like terms
0.306 - 0.0055 = 1609x
0.3005 = 1609x
x = 0.3005/1609
x = 1.8676 × 10⁻⁴
x ≅ 0.0002 M
No. of grams for the chlorophyll can be computed as follows:
recall that:
molar mass of chlorophyll = 893.5 g/mol
the volume = 25ml = (25/1000) L = 0.025 L
∴
In spinach solution, the no. of grams for the chlorophyll:
= (0.0002) mol/L × (893.5 g/mol) × (0.025) L
= 0.0044675 g
≅ 0.0045 g
In the spinach, the concentration of chlorophyll = no of grams of chlorophyll/ mass of the spinach
= 4.5 mg/0.1876 g
= 23.987 mg/g
≅ 24 mg/g
From the given information:
We are to make use of the spinach absorbance extract which is the corrected absorbance (y) = 0.306ChlorophyllChlorophyll is any member of the class of the green pigments involved in the photosynthesis process.
And also the trendline equation:
y = 1609x + 0.0055
where,
x = absorbance of the spinach extract.
so 0.306 = 1609x + 0.0055
collecting the like terms
0.306 - 0.0055 = 1609x
0.3005 = 1609x
x = 0.3005/1609
x = 1.8676 × 10⁻⁴
x ≅ 0.0002 M
2. No.of grams for the chlorophyll can be computed as follows:
recall that:
molar mass of chlorophyll = 893.5 g/mol
The volume = 25ml = (25/1000) L = 0.025 L
Therefore:
In spinach solution, the no. of grams for the chlorophyll:
= (0.0002) mol/L × (893.5 g/mol) × (0.025) L
= 0.0044675 g
≅ 0.0045 g
3. In the spinach, the concentration of chlorophyll = no of grams of chlorophyll/ mass of the spinach
= 4.5 mg/0.1876 g
= 23.987 mg/g
≅ 24 mg/g
Read more about chlorophyll here:
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a. Draw 2,3-dichloro octane.
b. Write the lewis structure for H20 molecule.
Answer:
a.draw 2,3 dicholoro octane
Explanation:
mag isip ka kung paano hehe
Sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are ________partially miscible . This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are__________
Answer: Sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are both partially miscible. This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are miscible.
Explanation:
When a substance (solute) dissolves partially in a solvent then it is known as partially miscible in the solvent. In such cases, a small amount of solute remains at the bottom of solution.
If a solute dissolves completely in solvent like water such that only one layer is seen in the solution then it means that the solute is miscible in solvent.
Thus, we can conclude that sugar is added to water and initially completely dissolves, but eventually solid sugar collects on the bottom of the container. Sugar and water are both partially miscible. This produces a dynamic equilibrium. Ethanol (a liquid) is added to water and only a single layer is observed no matter how much ethanol is added. Ethanol and water are miscible.
Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g) NH3(g)
(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.
Answer:
a) N2 (g) + H2 = 2 NH3
b) You have to state the mass of hydrogen
43.0 mL of 1.49 M perchloric acid is added to 14.0 mL of calcium hydroxide, and the resulting solution is found to be acidic.
29.1 mL of 0.498 M barium hydroxide is required to reach neutrality.
What is the molarity of the original calcium hydroxide solution?
Answer:
2.29 M
Explanation:
Equation of the reaction;
Ca(OH)2(aq) + 2HClO4(aq) → 2H2O(l) + Ca(ClO4)2(aq)
Concentration of acid CA = 1.49 M
Concentration of base CB= ????
Volume of acid VA= 43.0 ml
Volume of base VB= 14.0 ml
Number of moles of acid NA = 2 moles
Number of moles of base NB = 1 mole
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
CB= CAVANB/VBNA
CB= 1.49 × 43.0 × 1/14.0 × 2
CB= 2.29 M
11th grade chemistry question will mark brainliest
2.50 g of CO2 gas is confined in a rigid cylinder at a pressure
of 4.65 atm. If 0.42 g of gas is released from the cylinder,
what is the new pressure?
Answer:
3.88 atm
Explanation:
We'll begin by calculating the number of mole of CO₂ in each case. This can be obtained as follow:
For 2.50 g of CO₂:
Mass of CO₂ = 2.5 g
Molar mass of CO₂ = 12 + (2×16) = 44 g/mol
Mole of CO₂ =?
Mole = mass / molar mass
Mole of CO₂ = 2.5 / 44
Mole of CO₂ = 0.06 mole
For 0.42 g of CO₂:
Mass of CO₂ = 2.5 g
Molar mass of CO₂ = 44 g/mol
Mole of CO₂ =?
Mole = mass / molar mass
Mole of CO₂ = 0.42 / 44
Mole of CO₂ = 0.010 mole
Finally, we shall determine the new pressure. This can be obtained as follow:
Initial mole (n₁) = 0.06 mole
Initial pressure (P₁) = 4.65 atm
Final mole (n₂) = 0.06 – 0.010 = 0.05 mole
Final pressure (P₂) =?
NOTE: Temperature and volume is constant.
P₁ / n₁ = P₂ / n₂
4.65 / 0.06 = P₂ / 0.05
Cross multiply
0.06 × P₂ = 4.65 × 0.05
0.06 × P₂ = 0.2325
Divide both side by 0.06
P₂ = 0.2325 / 0.06
P₂ = 3.88 atm
Thus, the new pressure is 3.88 atm.
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Question 1
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Initial Knowledge Check
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Write the symbol for every chemical element that has atomic number less than 14 and atomic mass greater than 23.2 u.
Answer:
8 oxygen. 9 flourine. 10. Neon. 5 Boron
Which term can be used to describe the process in the reaction below? 2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
Answer:
Decomposition
Explanation:
If we look at the process;
2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
We can see that NaHCO3 was broken down into Na2CO3, H2O and CO2.
The breakdown of one compound to yield other chemical compounds is known as decomposition.
Hence the NaHCO3 was decomposed in the process above.
Why ethanol is used in pectin extraction
Explanation:
is responsible for interrupting the interaction between pectins and solvent molecules
g A sample of chlorine gas starting at 681 mm Hg is placed under a pressure of 991 mm Hg and reduced to a volume of 513.7 mL. What was the initial volume, in mL, of the chlorine gas container if the process was performed at constant temperature?
Answer:
747.5 mL
Explanation:
Assuming ideal behaviour, we can solve this problem by using Boyle's law, which states that at constant temperature:
P₁V₁ = P₂V₂Where in this case:
P₁ = 681 mm HgV₁ = ?P₂ = 991 mm HgV₂ = 513.7 mLWe input the data given by the problem:
681 mm Hg * V₁ = 991 mm Hg * 513.7 mLAnd solve for V₁:
V₁ = 747.5 mLdefine saturated and unsaturated fats
Answer:
unsaturated fats, which are liquid at room temperature,are different from saturated fat because they contain one or more double bonds and fewer hydrogen atoms on their carbon chain.
Many home barbeques are fueled with propane gas (C3H8)(C3H8). Part A What mass of carbon dioxide is produced upon the complete combustion of 27.9 LL of propane (the approximate contents of one 5-gallon tank)
Answer:
41264 g of CO₂
Explanation:
Combustion reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.
In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.
We have the volume of propane but we need moles of it, so we need to apply density.
Density = mass / volume so mass = density . volume.
Density of propane is: 493 g/L
Mass of propane is 493 g/L . 27.9L = 13754.7 g
We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles
According to reaction, 1 mol of propane can produce 3 moles of CO₂
Our 312.6 moles will produce 312.6 . 3 = 937.8 moles
We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g
What should be done if a spectrophotometer reports an absorbance that is too high? Select one: Reposition the cuvette in the spectrophotometer. Pour out half the volume of the sample. Restart the spectrophotometer and try again. Dilute the sample. g
Answer:
The sample should be diluted
Explanation:
According to Beer Lambert's law, the absorbance of a sample depends on the concentration of the sample.
Hence, if the concentration of the sample is very high, the spectrophotometer will also report a very high value of absorbance.
When this is the case, the sample should simply be diluted and the readings are taken again using the spectrophotometer.
Why the catalytic and optical properties of nanomaterial are different from bulk material
Answer:
The material properties of nanostructures are different from the bulk due to the high surface area over volume ratio and possible appearance of quantum effects at the nanoscale. ... Yu; they found that the structural distortions on the quantum dots depend both on the kind of dopant and on the size of the dots.
Explanation:
hope it helps
When a photon of red light hits metal X, an electron is ejected. Will an electron be ejected if a photon of yellow light hits metal X?
A. Yes
B. No
C. Can't tell
calculate the molarity in a 0.550 m solution of NaCl in water. Assume that the solution density is 1.03g/mol
Answer:
M=0.549M
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to perform this calculation by firstly assuming we have 1 kg of water as the solvent so that we have 0.550 moles of NaCl as well. Moreover, we realize we have 1000 grams of water and the correct mass of the solution can be calculated by converting 0.550 moles of NaCl to grams by using its molar mass:
[tex]m_{solute}=0.550mol*\frac{58.44 g}{1mol}= 32.14g\\\\m_{solution}=1000g+32.14g=1032.14g[/tex]
And subsequently, the volume in liters by using the density and the correct conversion factor:
[tex]V_{solution}=1032.14g*\frac{1mL}{1.03g} *\frac{1L}{1000mL} =1.002L[/tex]
Finally, the molarity will be:
[tex]M=\frac{0.550mol}{1.002L} =0.549M[/tex]
Regards!
What happens when Sulphur dioxide (so2) gas is passed through an acidified solution of hydrogen . sulfide (H₂S) gas :
Answer:
When SO
2
is passed through an acidified solution of H
2
S, sulphur is precipitated out according to the reaction.
2H
2
S+SO
2
→2H
2
O+3S
Draw bond-line formulas of all monochloro derivatives that might be formed when 1,1-dimethylcyclobutane is allowed to react with Cl2 under UV irradiation. For each structure, indicate, with an asterisk, any stereocenters that might be present.
Answer:
Attached below
Explanation:
Diagram of Bond-line formulas of all monochloro derivatives formed when 1,1-dimethylcyclobutane is allowed to react with c12 under UV
attached below
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to lowest. CoCl3, NH4Cl, Li2SO4
Answer:
NH4Cl > Li2SO4 > CoCl3
Explanation:
Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.
Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.
Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.
Give the following reaction: ammonium nitrate—> dinitrogen monoxide + water.
a.) Write a complete balanced chemical equation.
b.) Calculate the number of molecules of water produced by 11.2g of ammonium nitrate
Answer:
a) NH₄NO₃ ⇒ N₂O + 2 H₂O
b) 1.69 × 10²³ molecules
Explanation:
Step 1: Write the balanced equation
NH₄NO₃ ⇒ N₂O + 2 H₂O
Step 2: Convert 11.2 g of NH₄NO₃ to moles
The molar mass of NH₄NO₃ is 80.04 g/mol.
11.2 g × 1 mol/80.04 g = 0.140 mol
Step 3: Calculate the moles of H₂O produced
0.140 mol NH₄NO₃ × 2 mol H₂O/1 mol NH₄NO₃ = 0.280 mol H₂O
Step 4: Calculate the number of molecules in 0.280 moles of water
We will use Avogadro's number.
0.280 mol × 6.02 × 10²³ molecules/1 mol = 1.69 × 10²³ molecules
show in chemistry the fermentation of carbohydrate to form alcohol
hope it helps to everyone
The first law of thermodynamics defines chemical energy. defines entropy. is a statement of conservation of energy. provides a criterion for the spontaneity of a reaction.
Answer: The first law of thermodynamics is a statement of conservation of energy.
Explanation:
According to the first law of thermodynamics, heat provided to a system is actually the sum of internal energy and work done by the system or on the system.
Mathematically, [tex]\Delta Q = \Delta U + \Delta W[/tex]
The first law of thermodynamics also means that energy can neither be created nor it can be destroyed. Hence, energy is conserved.
Thus, we can conclude that the first law of thermodynamics is a statement of conservation of energy.
D
Question 2
1 pts
How many moles of carbon tetrachloride are present in
18.Og of CC14?
i think .117018
now to get to the 20charaeactrs minimum.
A 46.6-mgmg sample of boron reacts with oxygen to form 150 mgmg of the compound boron oxide. Part A What is the empirical formula of boron oxide
Answer:
B₂O₃
Explanation:
Step 1: Calculate the mass of oxygen in 150 mg of boron oxide
Of 150 mg of boron oxide, 46.6 mg belong to boron. The mass of oxygen is:
150 mg - 46.6 mg = 103.4 mg
Step 2: Calculate the percent by mass of each element
We will use the following expression.
%Element = mElement/mCompound × 100%
%B = 46.6 mg/150 mg × 100% = 31.1%
%O = 103.4 mg/150 mg × 100% = 68.9%
Step 3: Divide each percentage by the atomic mass of the element
B: 31.1/10.81 = 2.88
O: 68.9/16.00 = 4.31
Step 4: Divide both numbers by the smallest one (2.88)
B: 2.88/2.88 = 1
O: 4.31/2.88 ≈ 1.5
Step 5: Multiply both numbers by 2 so that they are integers
B: 1 × 2 = 2
O: 1.5 × 2 = 3
The empirical formula is B₂O₃.
Typhoon signals rise due to what? wind speed or wind strength or both?
A hypnotist's watch hanging from a chain swings back and forth every 0.98 s. What is the frequency (in Hz) of its oscillation?
Answer:
1.02 Hz
Explanation:
frequency= (1/t) = (1/0.98) = 1.02 hz
g Elimination reaction simulation, Kim had tried to prepare cyclohexene from cyclohexanol using HCl as a catalyst, but he got a weird side product cyclohexyl chloride. Cyclohexyl chloride is formed possibly from the side reaction called A. SN1 substitution reaction B. E2 reaction C. E1 reaction D. SN2 substitution reaction
Answer:
SN1 substitution reaction
Explanation:
cyclohexanol is a secondary alkanol. The mechanism of acid catalysed dehydration of cyclohexanol involves the protonation of the -OH group.
This is followed by loss of water which is a good leaving group. At this stage, a proton could be abstracted to yield cyclohexene by E1 mechanism or an SN1 substitution reaction may occur to yield Cyclohexyl chloride.
The both reactions are equally likely.
In the titration of 82.0 mL of 0.400 M HCOOH with 0.150 M LiOH, how many mL of LiOH are required to reach the equivalence point
Answer:
218.7 mL
Explanation:
The reaction that takes place is:
HCOOH + LiOH → LiCOOH + H₂OFirst we calculate how many HCOOH moles reacted, using the given volume and concentration:
82.0 mL * 0.400 M = 32.8 mmol HCOOHAs 1 HCOOH mol reacts with 1 LiOH mol, 32.8 mmoles of LiOH are needed to react with 32.8 mmoles of HCOOH.
Finally we calculate how many mL of a 0.150 M solution would contain 32.8 mmoles:
32.8 mmol / 0.150 M = 218.7 mLPredict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.)
______________ → BaBr2 + H2O
Answer:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
Explanation:
We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
