Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
A student dropped a pea size amount of K2CO3 into a solution of HCl(aq). He observed the formation of gas bubbles and collected the gas into another test tube. The student performed a splint test and observed that the splint was extingished when he placed the splint into the test tube of the gas. What can be said about the results of this students experiment?
a. The student completed the experiment correctly and there were no errors in the experiment.
b. The experiment was performed incorrectly. K2CO3 doesn't react with HCl. Therefore, the student picked up the wrong compound when conducting the experiment.
c. The student performed the splint test incorrectly. He should of observed the splint flare up when the splint was placed in the test tube.
d. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
Answer:
The student completed the experiment correctly and there were no errors in the experiment.
Explanation:
When a pea size amount of K2CO3 is dropped into a solution of HCl, the following reaction occurs;
K2CO3(s) + 2HCl(aq) ----> 2KCl(aq) + CO2(g) + H2O(l)
The gas CO2 does not support burning hence, when the student performed a splint test and observed that the splint was extinguished when he placed the splint into the test tube of the gas.
Hence, the experiment was properly conducted and the student completed the experiment correctly and there were no errors in the experiment.
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.
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.
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
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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
show in chemistry the fermentation of carbohydrate to form alcohol
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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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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 mLThe chemical formula is different from the empirical formula in
Answer:be careful and relax
Explanation:
Answer:
Hahaha be careful and relax
Predict 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
Typhoon signals rise due to what? wind speed or wind strength or both?
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
Why ethanol is used in pectin extraction
Explanation:
is responsible for interrupting the interaction between pectins and solvent molecules
Water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the bottom.
ANSWER please
True
False
Answer:
False
Explanation:
An object with a density less than the density of water will float.
And an object that is denser than water (more than 1.00g/mL) will sink when placed in water.
How do you predict the geometrical shape of NH3 on VSEPR model
Answer:
NH3 Ammonia
Explanation:
Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.
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.
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.
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.
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
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
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.
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.
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.
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 mLGive 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
Identify the intermolecular attractions for dimethyl ether and for ethyl alcohol. Which molecule is expected to be more soluble in water? Explain.
Answer:
See explanation
Explanation:
All molecules possess the London dispersion forces. However London dispersion forces is the only kind of intermolecular interaction that exists in nonpolar substances.
So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.
As for ethyl alcohol, the molecule is polar due to the presence of polar O-H bond. In addition to London dispersion forces, dipole-dipole interactions and specifically hydrogen bonding also occurs between the molecules.
Because ethyl alcohol is polar, it is more soluble in water than dimethyl ether.
Observe as equações e de acordo com Brönsted-Lowry, os compostos destacados são, respectivamente
answer
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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
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.
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
