The electron releases energy in the form of light.
Hope this helps!
3. If you need to produce 85 g of CO2, how many grams of: (these are 3 problems starting with thea. C3H8, do you need?same amount:b. O2, do you need?c. H2O will also be made?
1) First let's write the equation. It is a combustion reaction, so:
C₃H₈ + O₂ ---> CO₂ + H₂O
and balance the equation (same number of atoms of each element on both sides of the equation):
C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O
Reactant side:
C - 3
H - 8
O - 10
Product side:
C - 3
O - 10
H - 8
2) Now let's transform 85 grams of CO₂ into mole. For this, we use the following equation:
mole = mass/molar mass
Molar mass of CO₂ is: (1×12) + (2×16) = 44 g/mol
mole = 85/44
mole = 1.9 mol of CO₂
3) Now we use the proportion of the balanced equation:
1 mol of C₃H₈ ---- 3 mol of CO₂
x mol of C₃H₈ ----- 1.9 mol of CO₂
x = 0.6 mol of C₃H₈
4) Now we transform mole of C₃H₈ into grams using its molar mass.
molar mass of C₃H₈ is: (3×12) + (8×1) = 44 g/mol
mass = mole × molar mass
mass = 0.6 × 44
mass of C₃H₈ = 28 g
Answer: a) mass of C₃H₈ = 28 g
For alternative b we follow the same process starting from step 3:
3)Now we use the proportion of the balanced equation:
5 mol of O₂ ---- 3 mol of CO₂
x mol of O₂ ----- 1.9 mol of CO₂
x = 3.16 mol of O₂
4) Now we transform mole of O₂ into grams using its molar mass.
molar mass of O₂ is: (2×16) = 32 g/mol
mass = mole × molar mass
mass = 3.16 × 32
mass of O₂ = 101 g
Answer: b) mass of O₂ = 101 g
For alternative c we follow the same process starting from step 3:
3) Now we use the proportion of the balanced equation:
4 mol of H₂O ---- 3 mol of CO₂
x mol of H₂O ----- 1.9 mol of CO₂
x = 0.84 mol of H₂O
4) Now we transform mole of H₂O into grams using its molar mass.
molar mass of H₂O is: (2×1) + (1×16) = 18 g/mol
mass = mole × molar mass
mass = 0.84 × 18
mass of H₂O = 15 g
Answer: c) mass of H₂O = 15 g
A reaction experimentally yields 15.68 g of a product. What is the percent yield if the theoretical yield is 18.81 g?
Answer
The percent yield = 83.36%
Explanation
Given:
Experimental yield = actual yield = 15.68 g
Theoretical yield = 18.81 g
What to find:
The percent yield for the reaction.
Step-by-step solution:
The percent yield for the reaction can be calculated using the formula below:
[tex]\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{15.68\text{ }g}{18.81\text{ }g}\times100\% \\ \\ Percent\text{ }yield=83.36\% \end{gathered}[/tex]Hence, the percent yield for the reaction is 83.36%
Substance Density (grams/cm3)Chloroform - 1.5Ebony wood - 1.2Mahogany wood - 0.85Oil - 0.9Water - 1.025.Since volume = mass/density, a 1,700 gram beam of mahogany wood has a volume of...Volume = Mass / DensitySelect one:a. 500 cm3b. 1,445 cm3c. 1,785 cm3d. 2,000 cm3
As the question gave us the formula in which we have to use to calculate the volume of this type of wood:
V = m/d
We have:
m = 1700 grams
d = 0.85
Now we add these values into the formula:
V = 1700/0.85
V = 2000 cm3, letter D
Describe global influences on local
weather.
54000 mL isO 54 m3O 5400 cm3O 0.054m3O 540 m3
mL and cm³ have a 1 to 1 conversion, so we have:
[tex]54000mL=54000cm^{3}[/tex]But we don't have this option, so we will need to find another.
The other are in m³, so we can use the conversion from cm to m to get this:
[tex]\begin{gathered} 1m=100cm \\ (1m)^{3}=(100cm)^{3} \\ 1m^{3}=1000000cm^{3} \\ 1cm^{3}=\frac{1}{1000000}m^{3} \end{gathered}[/tex]So, we can apply this to what we have:
[tex]54000cm^3=54000\cdot\frac{1}{1000000}m^3=0.054m^{3}[/tex]We have an option with 0.054m³, so the correc alternative is 0.054 m³.
Find the element that is oxidized and the one that is reduced Si + 2 F2 --> SiF4
Answer
The element that is oxidized is Si and the one that is reduced is F₂.
Explanation
Si + 2F₂ → SiF₄
The given reaction is an oxidation-reduction (redox) reaction:
Oxidation: Si → Si⁴⁺ + 4e⁻
Reduction: 2F₂ + 4e⁻ → 4F⁻
Si is a reducing agent, and F₂ is an oxidizing agent.
An oxidizing agent gains electrons and is reduced in a chemical reaction.
A reducing agent loses electrons and is oxidized in a chemical reaction.
Therefore, the element that is oxidized is Si and the one that is reduced is F₂.
the atomic masses of the two stable isotopes of beam
10.81amu
Explanations:In order to get the average atomic mass of an element, we need the following parameters:
• Natural Abundance (NA),: The percentage of atoms for an element that is a specific isotope.
• Mass (m) ,of each isotope
For the given element (Boron-10 and Boron-11), the natural abundances are 19.78% and 80.22% respectively.
The atomic masses of Boron-10 and Boron-11 are 10.0129amu and 11.0093amu respectively
The formula for calculating the average atomic mass of the element is expressed as:
[tex]AAM=(NA_a\times m_a)+(NA_b\times m_b)[/tex]Substitute the given parameters into the formula to have:
[tex]A\mathrm{}A\mathrm{}M=(0.1978\times10.0129)+(0.8022\times11.0093)[/tex]Simplify the resulting expression to have:
[tex]\begin{gathered} A\mathrm{}A\mathrm{}M=1.98055162+8.83166046 \\ A\mathrm{}A\mathrm{}M=10.81221208 \\ A\mathrm{}A\mathrm{}M\approx10.81amu \end{gathered}[/tex]Therefore the average atomic mass of Boron is 10.81amu to two decimal places.
Write formulas or names as appropriate for each of the following ionic compounds. 1. Magnesium nitride 6. SrI2 2. Lithium oxide 7. Ba3(PO4)2 3. Aluminum sulfite 8. (NH4)2O 4. Copper(II) bicarbonate 9. Fe(ClO)3 5. Sodium nitrate 10. ZnCrO4
We have the following formulas for the given compounds:
1. Magnesium nitride ---> Mg3N2
2. Lithium oxide ---> Li2O
3. Aluminum sulfite ---> Al2(SO4)3
4. Copper (II) bicarbonate ---> Cu(HCO3)2
5. Sodium nitrate ---> NaNO3
For the given formulas we have the following names:
6. SrI2 ---> Strontium iodide
7. Ba3(PO4)2 --->Barium phosphate
8. (NH4)2O ---> Ammonium oxide
9. Fe(ClO)3 ---> Iron(III) hypochlorite
10. ZnCrO4 ---> zinc chromate
How many atoms of O are there in 7.00 g FeSO4 ?
Answer:
1.11 x 10²³ atoms O
Explanation:
To find the number of oxygen atoms in FeSO₄, you need to (1) convert FeSO₄ from grams to moles (using the molar mass), then (2) convert moles FeSO₄ to moles O (using the mole-to-mole ratio of FeSO₄), and then (3) convert moles O to atoms O (using Avogadro's Number). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs like the given number (7.00 = 3 sig figs).
Atomic Mass (Fe): 55.845 g/mol
Atomic Mass (S): 32.065 g/mol
Atomic Mass (O): 15.999 g/mol
Molar Mass (FeSO₄): 55.845 g/mol + 32.065 g/mol + 4(15.999 g/mol)
Molar Mass (FeSO₄): 151.906 g/mol
1 mole FeSO₄: 1 mole Fe, 2 mole S, 4 moles O
Avogadro's Number:
6.022 x 10²³ atoms = 1 mole
7.00 g FeSO₄ 1 mole 4 moles O 6.022 x 10²³ atoms
----------------------- x ------------------- x ----------------------- x ----------------------------
151.906 g 1 mole FeSO₄ 1 mole
= 1.11 x 10²³ atoms O
How many grams of H2 are required to completely convert 80g of Fe2O3?
Answer
3.0 grams H₂ is required.Explanation
Given:
Mass of Fe2O3 = 80 g
Equation:
What to find:
The grams of H2 required to completely convert 80g of Fe2O3.
Step-by-step solution:
From the equation of reaction;
3 moles of H2 completely react with 1 mole of Fe2O3
Note: Molar mass of H2 is 2.016 grams per mole and Molar mass of Fe2O3 is 159.69 g/mol
This implies; (3 x 2.016 g) = 6.048 grams H2 completely react with 159.69 grams Fe2O3.
Therefore, x grams H2 will completely convert 80 grams Fe2O3.
Cross multiply and divide both sides by 159.69 grams Fe2O3.
x grams H2 is now equal to
[tex]x=\frac{80\text{ }g\text{ }Fe_2O_3}{159.69\text{ }g\text{ }Fe_2O_3}\times6.048\text{ }g\text{ }H_2=3.0298\approx3.0\text{ }grams\text{ }H_2[/tex]Therefore the grams of H2 required to completely convert 80g of Fe2O3 is 3.0 grams
A gas occupying 0.6 L at 1.70 atm expands to 0.9 L. What is the new pressure assuming temperature remains constant?
Answer:
1.13 atmExplanation:
The new pressure can be found by using the formula
P1V1 = P2V2
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the new pressure P2 we make P2 the subject
We have
[tex]p_2 = \frac{p1v1}{v2} \\ [/tex]
P1 = 1.7 atm
V1 = 0.6 L
V2 = 0.9 L
We have
[tex]p_2 = \frac{1.7 \times 0.6}{0.9} = 1.13333...\\ [/tex]
We have the final answer as
1.13 atmHope this helps you
The following lists consists of ionic compounds EXCEPT
barium hydroxide, zinc carbonate, ammonium sulfate
calcium chloride, carbon disulfide, magnesium nitrate
sodium sulfate, copper(II) oxide, potassium nitride
aluminium sulfide, sodium sulfite, calcium fluoride
The following lists consists of ionic compounds except carbon disulfide (CS₂).
Barium hydroxide , Ba(OH)₂ is an ionic compound.
zinc carbonate, ZnCO₃ is an ionic compound.
ammonium sulfate , (NH₄)₂SO₄ is an ionic compound.
calcium chloride, CaCl₂ is an ionic compound.
carbon disulfide, CS₂ is not an ionic compound. In carbon disulfide both the elements are non metallic elements. The bond formed between atoms are by sharing of electron known as covalent bond due to very little difference in electronegativity.
magnesium nitrate, Mg(NO₃)₂ is an ionic compound.
sodium sulfate, Na₂SO₄ is an ionic compound.
copper(II) oxide, CuO is an ionic compound.
potassium nitride KNO₃ is an ionic compound.
aluminium sulfide, Al₂S₃ is an ionic compound.
sodium sulfite, Na₂S is an ionic compound.
calcium fluoride, CaF₂ is an ionic compound.
Thus, The following lists consists of ionic compounds except carbon disulfide (CS₂).
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A student finds a piece of metal and finds its mass to be 750 g. Through water displacement thestudent determines the volume to be 65.8 cm". Which metal does the student have?
Metals normally have different densities, so we can try to determine which metal is this by its density.
Assuming it is pure, the density of the metal is its mass divided by its volume:
[tex]\begin{gathered} \rho=\frac{m}{V} \\ \rho=\frac{750g}{65.8cm^3} \\ \rho\approx11.4g/cm^3 \end{gathered}[/tex]Now, we need to look for some table with densities of various metals and see which one has the density we found, 11.4 g/cm³.
Since we don't have one, we can look for one. In it, we can see that the only metal with this density is lead.
So, the metal in the question should be lead.
Ascorbic acid (vitamin C) is important in many metabolic reactions in the body, including the synthesis of collagen and prevention of scurvy. Given that themass percent composition of ascorbic acid is 40.9% C, 4.58% H, and 54.5% O, determine the empirical formula of ascorbic acid. Show all your work
Empirical formula:
Step 1
Information already provided
The mass percent composition:
40.9 % C
4.58 % H
54.5 % O
Information needed: from the periodic table
For C) 1 mol = 12.01 g
For H) 1 mol = 1.008 g
For O) 1 mol = 15.99 g
---------------------------
Step 2
A sample of 100 g is assumed, so:
40.9 % C => 40.9 g C
4.58 % H => 4.58 g H
54.5 % O => 54.5 g O
--------------------------
Step 3
Convert mass into moles:
40.9 g C x (1 mol/12.01 g) = 3.40 moles C
4.58 g H x (1 mol/1.008 g) = 4.54 moles H
54.5 g O x (1 mol/15.99 g) = 3.40 moles O
------------------------
Step 4
All moles calculated in step 3 need to be divided by the smallest one.
3.40 moles C/3.40 moles = 1
4.54 moles H/3.40 moles = 1.33
3.40 moles O/3.40 moles = 1
-----------------------
Step 5
Integer numbers are needed, so let's multiply by 3 all of them in step 4
Therefore,
For C) 3
For H) 3.99 = 4 approx.
For O) 3
All these numbers calculated will be the subindexes in ascorbic acid
Answer:
Empirical formula: C3H4O3
Determine the percent composition of hydrogen for the following: NaHCO3
By definition, the percent composition of an atom in a compound is its mass percentage in the formula.
That is, if we have 1 mol of NaHCO₃, we have also 1 mol of H (because there is only on H for each molecule).
So, we calculate the mass of this 1 mol of NaHCO₃ and the mass of 1 mol of H and calculate the percentage.
In equations, we want the following:
[tex]C_H=\frac{m_H}{m_{NaHCO_{3}}}[/tex]Since these are ratios, we doesn't matter if we talk about 1, 2 or any number of moles, but 1 mol is easier because the molecular and atomi masses are for 1 mol.
The molecular mass of NaHCO₃ is:
[tex]\begin{gathered} M_{NaHCO_3}=M_{Na}+M_H+M_C+3\cdot M_O \\ M_{NaHCO_3}\approx(22.990+1.008+12.011+3\cdot15.999)g/mol \\ M_{NaHCO_3}\approx84.006g/mol \end{gathered}[/tex]Which means that we have approximately 84.006 grams of NaHCO₃ in 1 mol of it.
The atomic mass of H is:
[tex]M_H\approx1.008g/mol[/tex]Which means that we have approximately 1.008 grams of H in 1 mol of it.
Now, we can take the percentage of mass of H:
[tex]C_H\approx\frac{1.008g_{}}{84.006g}\cdot100\%\approx1.20\%[/tex]So, the percentage composition of H in NaHCO₃ is approximately 1.20%.
Write a balanced equation for the decomposition reaction that occurred in Experiment 2. Include physical states.
What is the largest possible electronegativity difference for a bond to be covalent?A.0.5B.1.7C.0.0D.1.0
Answer
B. 1.7
Explanation
As a rule, an electronegativity difference of 2 or more on the Pauling scale between atoms leads to the formation of an ionic bond. A difference of less than 2 between atoms leads to covalent bond formation.
Therefore, the largest possible electronegativity difference for a bond to be covalent is 1.7
17 was found from determination in a mass spectrometer that an element X has three Isotopes whose mass are & 19.19,20.99 and 21.99 respectively The abundance of these I sotopes are 90.92% 0.25% $8.83% respectively Calculate the relative atomic mass.
The relative atomic mass of the given element is 40.372 amu.
What is relative atomic mass?The relative atomic mass of an element is considered as the sum of the isotopes masses each multiplied by the percentage which is found in nature.
The formula which is used to calculate the relative atomic mass is
Relative atomic mass = sum of all atomic masses of isotopes × fractional abundance
Given,
Mass of isotopes 1 = 19.19 amu
Mass of isotopes 2 = 20.99 amu
Mass of isotopes 3 = 21.99 amu
Fractional abundance of isotope 1 = 0.9092
Fractional abundance of isotope 2 = 0.0025
Fractional abundance of isotope 3 = 0.0883
By substituting all the values, we get
[( 19.19 × 0.9092) + (20.99 × 0.0025) + (21.99 × 0.883)]
= 17.447 + 0.052 + 22.873
= 40.372 amu.
Thus, we concluded that the relative atomic mass of the given element is 40.372 amu.
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At 302 K , to what pressure can the carbon dioxide in the cartridge inflate a 3.05 L mountain bike tire? (Note that the gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi .)
Ideal gas law is valid only for ideal gas not for vanderwaal gas. Therefore the carbon dioxide in the cartridge inflate a 3.05 L mountain bike tire to 104.3psi pressure.
What is ideal gas equation?Ideal gas equation is the mathematical expression that relates pressure volume and temperature.
Mathematically,
PV=nRT
where,
P = pressure
V= volume=3.05 L
n =number of moles=1mole(assumed as it is not given in question)
T =temperature = 302 K
R = Gas constant = 0.0821 L.atm/K.mol
P × 3.05 L =1 mole× 0.0821 L.atm/K.mol × 302 K
P =8.12atm
1 atm = 14.7 psi
Hence Pressure in psi = 8.12atm×14.7 psi = 119.49 psi
Pressure by the gas= Total pressure - Atmospheric pressure = 119.49 - 14.7 psi = 104.3psi
Therefore the carbon dioxide in the cartridge inflate a 3.79 L mountain bike tire to 104.3psi pressure.
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One day in lab, while adding a gnarled root to a dark liquid bubbling in an iron cauldron, your friend Leila (an expert chemist) says this:
"Group 1A metal hydrides react with water to produce hydroxides and hydrogen gas."
Using Leila's statement, and what you already know about chemistry, predict the products of the following reaction.
Be sure your chemical equation is balanced!
KH(s)+ H2O(l)=
The balanced equation of the reaction of the metal hydride, KH, and water is given below:
KH (s) + H₂O(l) ---> KOH + H₂ (g) What are metal hydrides?Metal hydrides are compounds that are composed of a reactive metal chemically combined with hydrogen.
The hydrogen atom is less electropositive than the metal and will accept electrons from the metal to form the negative hydride ion.
For example, Group 1A metal hydrides react with water to produce hydroxides and hydrogen gas.
Considering the above property, the given reaction will be:
KH (s) + H₂O(l) ---> KOH + H₂ (g)
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List two components of rocket fuel?
A component that many space agencies use is liquid hydrogen, which can be abbreviated as LH2.
And another fuel very used in rockets is kerosene which is a hydrocarbons mixture of other compounds that fuel.
Answer:
Most liquid chemical rockets use two separate propellants: a fuel and an oxidizer. Typical fuels include kerosene, alcohol, hydrazine and its derivatives, and liquid hydrogen. Many others have been tested and used. Oxidizers include nitric acid, nitrogen tetroxide, liquid oxygen, and liquid fluorine.
Explanation:
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A silver ring is composed of 1.81×1023 atoms. Calculate the mass of the ring in grams.
Mole equation: [tex]n=\dfrac{m}{MM}[/tex]
n = number of moles (mol)m = mass (g)MM = molar mass (g/mol)To find mass given number of atoms:
Divide number of atoms by the number of atoms in the chemical formula ⇒ find number of moleculesDivide number of molecules by Avogadro's number ([tex]6.02*10^{23}[/tex]) ⇒ find number of moles (n)Solve for m using moles equationSolving the QuestionWe're given:
Ag (silver)Atoms = [tex]1.81*10^{23}[/tex] atomsm = ?In the chemical formula, which is Ag, there is only 1 atom.
Divide [tex]1.81*10^{23}[/tex] atoms by 1 atom to get the number of molecules in the silver ring:
[tex]1.81*10^{23}\div 1\\=1.81*10^{23}[/tex]
Therefore, there are [tex]1.81*10^{23}[/tex] molecules in the silver ring.
Now, divide [tex]1.81*10^{23}[/tex] molecules by Avogadro's number to find n:
[tex]\dfrac{1.81*10^{23}}{6.02*10^{23}}\\\\=\dfrac{1.81}{6.02}\\\\=\dfrac{1.81}{6.02}\\\\= 0.30066[/tex]
Therefore, the sample has 0.30066 mol.
Finally, solve for the mass using the moles equation:
[tex]n=\dfrac{m}{MM}[/tex]
⇒ Rearrange the equation:
[tex]m=n*MM[/tex]
⇒ MM of Ag = 107.87 g/mol
⇒ Plug in given information:
[tex]m=0.30066* 107.87\\m=32.4[/tex]
Therefore, the mass of the ring is 32.4 g.
Answer
32.4 g
The value of AH rxn for the following reaction is -72 kJ. How many kJ of heat is released when 0.989 g of HBr (80.91 g/mol) is formed? H2 (g) + Br2 (g) -› 2 HBr (gram). A. -144B. -72 C. -0.44 D. -36
Answer:
[tex]C\text{ : -0.44 KJ}[/tex]Explanation:
Here, we want to get the amount of heat released in KJ
From the change in enthalpy given and the equation of reaction, we know that 2 moles of HBr would lead to that amount of heat
Now, let us get the actual amount of heat released
We need to get the actual number of moles of HBr produced
Mathematically, we can calculate that by dividing the mass of HBr by its molar mass
We have that as:
[tex]\frac{0.989}{80.91}\text{ = 0.0122 mol}[/tex]From the reaction information:
-72 KJ was released by 2 moles
x KJ would be released by 0.0122 mol
To get the value of x, we have it that:
[tex]\begin{gathered} x\text{ }\times2\text{ = 0.0122 }\times\text{ \lparen-72\rparen} \\ \\ x\text{ = -36 }\times\text{ \lparen0.0122\rparen = -0.44 KJ} \end{gathered}[/tex]17. Which of the following represents a formula for a chemical compound?A. CB. KOHC. O
KOH. Option B is correct
Explanations:A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.
The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)
What impact did cargo ship refrigeration systems have on the banana industry? Refrigeration slows the rate at which food is being spoiled, refrigeration does not affect speed at which the ship moves, refrigeration does not cause ripening, refrigeration does not control the sugar content of bananas?
Answer
Refrigeration slows the rate at which food is being spoiled
Explanation
One of the importance of storing foods at cold temperatures (refrigeration) is to slow the growth of microorganisms, thereby limiting food poisoning while preserving food's nutritional qualities and good taste.
Therefore, the impact the cargo ship refrigeration systems have on the banana industry is:
Refrigeration slows the rate at which food is being spoiled
Lithium nitride is an ionic compound. Draw a diagram which shows its formula, the charges on the ions and the arrangement of the valency (outer shell) electrons around the negative ion. Use x for an electron from a lithium atom and o for an electron from a nitrogen atom.
Lithium nitride is an ionic compound and the chemical formula of lithium nitride is Li₃N and the charges on the ions is Li⁺¹N³⁻ 3 valence electron in lithium and 5 valence electron in nitrogen
Ionic compound is the ion compound these ions are atom that gain or lose electron resulting in a net positive and negative charges and lithium nitride is an ionic compound and in that Li⁺¹N³⁻ is the ion and the arrangement of the valency (outer shell) electrons are 3 valence electron in lithium and 5 valence electron in nitrogen
X = 3 electron
O = 7 electron
And that's why lithium nitride has 10 electron
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calculate the theoretical and % yield of copper when 0.500 g of Cu was used and 0.350 g were recovered at the end of the experiment. complete solution
The theoretical percentage yield of copper is 70%
This is further explained below.
What is a yield?Generally, the equation for yield is mathematically given as
[tex]yield =\frac{\text { mote no. of product }}{\text { mote no. of reactant }} \times 100$[/tex]
Here, Based on Theoretical Yield
[tex]\begin{aligned} \text { mole rio. of reactant } &=-\frac{0.5}{63.546} \text { mole } \\ \end{aligned}$[/tex]
Based on the Actual yield
mole rio. of Product=0.35/63.546 mole
(Here atomic man of Cu=63.546g/mol )
Therefore, we apply the initially stated equation for %yield
So, %yield =[tex]\frac{\frac{0.35}{63.546}}{\frac{0.5}{(63.546}} \times 100 \%$[/tex]
=[tex]\frac{0.35}{0.5} \times 100 \%\\[/tex]
=0.7 *100%
=70%
In conclusion, the percentage yield is 70%
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Classify CH3CH2NH2 as astrong base or a weak base.Strong BaseWeak Base
Answer:
CH3CH2NH2 is a weak base.
Explanation:
CH3CH2NH2 is a weak base since its Kb is small, and thus it partially dissociates.
The pH of a basic solution is 8.13. What is [OH⁻]?
The [OH⁻] of the solution with pH of 8.13 is 1.35 * 10-6 M
pH is the measurement of the acidity or basicity of a compound by measuring [H⁻] ions in the solution. It ranges from 0 to 14 with acidic range from 0 – 7 and basic range from 7-14.
pOH is the measurement of the acidity or basicity of a compound by measuring [OH⁻] ions in the solution.
Thus, pH + pOH = 14
pOH = 14 - pH = 14 – 8.13 = 5.87
pOH = - log ([OH⁻])
- log ([OH⁻]) = 5.87
Log ([OH⁻]) = - 5.87
[OH⁻] = 10 ^ - 5.87 = 0.00000134896
[OH⁻] = 1.35 * 10⁻⁶ M
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Name three different forms of mixture
Answer:
Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions and colloids. The components of a mixture retain their own physical properties.
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Explanation: