Answer:
98.9%
Explanation:
2 moles of I₂ are required for each mole of Mg, so the reaction is limited by the available I₂. The 3.56 moles of I₂ should react with 1.78 moles of Mg to produce 1.78 moles of MgI₂. Instead, we get 1.76 moles of MgI₂.
The yield is 1.76/1.78 × 100% ≈ 98.876%
The yield is 98.9% of the quantity expected based on available reactants.
To determine the concentration of a sample of calcium hydroxide, 1.45M HCl is added drop-wise using a burst. Write the balanced net ionic equation for the acid-base reaction.
Answer:
H^+(aq) + OH^-(aq) —> H2O(l)
Explanation:
We'll begin by writing the balanced equation for the reaction.
2HCl(aq) + Ca(OH)2(aq) —> CaCl2(aq) + 2H2O(l)
Ca(OH)2 is a strong base and will dissociates as follow:
Ca(OH)2(aq) —> Ca^2+(aq) + 2OH^-(aq)
HCl is a strong acid and will dissociates as follow:
HCl(aq) —> H^+(aq) + Cl^-(aq)
Thus, In solution a double displacement reaction occurs as shown below:
2H^+(aq) + 2Cl^-(aq) + Ca^2+(aq) + 2OH^-(aq) —> Ca^2+(aq) + 2Cl^-(aq) + 2H2O(l)
To get the net ionic equation, cancel out Ca^2+ and 2Cl^-
2H^+(aq) + 2OH^-(aq) —> 2H2O(l)
H^+(aq) + OH^-(aq) —> H2O(l)
What kind of solid is crystalline boron (B)?
A. lonic solid
B. Metallic solid
C. Molecular solid
D. Network solid
Answer:
D
Explanation:
gr. 2.3 at 25°C; valence +3. Boron is a nonmetallic element existing as a dark brown to black amorphous powder or as an extremely hard, usually jet-black to silver-gray, brittle, lustrous, metallike crystalline solid
it is a network solid, a lattice of many covalent bonds (like diamond, except that it is black rather than transparent).
Network solid kind of solid is crystalline boron (B). Hence, option D is correct.
What is Network solid?A network solid is a solid where all the atoms are covalently bonded in a continuous network.
Boron is a nonmetallic element existing as a dark brown to black amorphous powder or as an extremely hard, usually jet-black to silver-grey, brittle, lustrous, metallike crystalline solid
It is a network solid, a lattice of many covalent bonds (like a diamond, except that it is black rather than transparent).
Hence, option D is correct.
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Suppose that a wave has a period of 0.03 seconds what is its frequency be sure to show the steps for your work
Answer:
f = 33.34 Hz
Explanation:
A wave has a period of 0.03 seconds. It is required to find the frequency of a wave. The relation between time period and frequency is inverse. The time period of a wave is given by :
T = 1/f, f = frequency of wave
[tex]f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.03}\\\\f=33.34\ Hz[/tex]
So, the frequency of the wave is 33.34 Hz.
Each of the insoluble salts below are put into 0.10 M hydrobromic acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution ?
a. Calcium sulfite
b. Calcium fluoride
c. Silver bromide
Answer:
A. Solubility of calcium sulfite increases
B. Solubility of calcium fluoride increases
C. Solubility of Silver bromide decreases
Explanation:
The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.
Case 1. Calcium sulfite
The dissociation reaction of CaSO3 is given below:
CaSO3 ----> Ca²+ + SO3²-
SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;
H+ + SO3²- ---> HSO3-
This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.
Case 2. Calcium fluoride
The dissociation reaction of calcium fluoride (CaF2) is shown below.
CaF2 ----> Ca²+ + 2F-
Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:
H+ + F- ---->. HF
Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.
Case 3: Silver bromide
The dissociation reaction of AgBr is as follows:
AgBr ----> Ag+ + Br-
The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.
The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.
Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.
Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.
However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.
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Coefficient of balanced equation: __Fe + ___020) — _Fe_036)
Answer:
- Four for iron, three for oxygen and 2 for iron (III) oxide:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
Explanation:
Hello,
In this case, the oxidation of iron is a widely acknowledged reaction occurring in ships and other machines exposed to the air or highly oxidizing medias. Thus, by the effect of oxygen, iron undergoes oxidation typically to iron (III) oxide:
[tex]Fe+O_2\rightarrow Fe_2O_3[/tex]
Nonetheless, the law of conservation of mass must be respected, therefore the coefficients balancing the reaction are four for iron, three for oxygen and 2 for iron (III) oxide:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
Best regards.
carbon dioxide is a non-polar molecule true or false
Answer:
True
Explanation:
Due to the arrangement of the molecule, a carbon dioxide molecule is non-polar.
Trans-4-hexen-3-ol can be synthesized starting from acetaldehyde. One of the key reagents is ethyl grignard.
1. Synthesize ethyl grignard from acetaldehyde in the steps below using the reagents provided.
2. Synthesize (trans)-4-hexen-3-ol from acetaldehyde.
find the given attachment
How many grams of 02 are needed to react with 7.50g of ethanol
How do the particles in plasmas compare with the particles in solids?
Answer:
Plasmas and solids are both made up of cation-anion pairs. Solids and plasmas are both made up of electrons and cations. Solids are made up of cation-anion pairs, but plasmas are not.
Answer:
Solids are made up of cation-anion pairs, but plasmas are not.
Explanation:
When vinylcyclohexane is treated with in dichloromethane, the major product is (2-bromo ethylidene)cyclohexane . Account for the formation of this product by drawing the structure of the most stable radical intermediate. Include all valence lone pairs in your answer. Include all valence radical electrons in your answer.
Answer:
Explanation:
Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.
SOo, when vinylcyclohexane is treated with NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .
This stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.
The diagram emphasizing more on the above explanation can be seen in the attached image below
In Chapter 4, we will learn that single bonds experience free rotation at room temperature, while double bonds do not. Consider the two C-N bonds in the structure. One of these bonds exhibits free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. Identify the C-N bond with restricted rotation, and justify your answer by drawing resonance structures.
Answer:
Explanation:
The main objective here is to draw a diagram of an heterocyclic compound containing two C-N bonds in the structure. One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. After that ; we will identify the C-N bond with restricted rotation, and also justify our answer by drawing resonance structures.
So; the first image below shows the structure of the heterocyclic compound containing two C-N bonds in the structure with One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. From the first diagram. the squared area indicates the C-N bond that exhibits restricted rotation.
The amide bonds in the C-N bonds offers the resonance characteristics and thus exhibits restricted rotation. The resonance is shown in the second image below
Which of the following is evidence for a physical change? A) burning B) fizzing C) evaporating D) rusting
Answer:c
Explanation: rusting, burning and fuzzing are all examples of chemical reactions/changes.
Three different students determined the density of a metal object. Here are their results: 15.12 g/mL, 15.09 g/mL, and 15.12 g/mL. The actual density of the object was 14.41 g/mL. Calculate the percent error. Make sure to include units with your answer, units are %.
Answer:
The correct answers are 4.93 %, 4.72 % and 4.93 %.
Explanation:
Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,
% error = (actual value - calculated value) / actual value * 100
By 1st student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41 - 15.12) / 14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,
% error = (14.41-15.09)/14.41 * 100
= 0.68/14.41 * 100
= 4.72 %
By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41-15.12)/14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later?
Answer:
below
Explanation:
28th 10;24 am
If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later at 12A.M.
How much time changes between Moon rises from one day to the next?This movement is from the Moon's orbit, which takes 27 days, 7 hours and 43 minutes to go full circle. It causes the Moon to move 12–13 degrees east every day. This shift means Earth has to rotate a little longer to bring the Moon into view, which is why moonrise is about 50 minutes later each day.
So knowing that moonrise is about 50 minutes later each day, we have:
[tex]7+50 minutes = 7:50\\8:40\\9:30\\10:20\\11:10\\12:00 A.M[/tex]
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MnCl₂ (aq) + (NH₄)₂CO₃ (aq) --> MnCO₃ (s) + 2 NH₄Cl (aq)
this equation is balanced.
Answer:
Nothing to do, is already balanced.
Explanation:
MnCl₂ (aq) + (NH₄)₂CO₃ (aq) --> MnCO₃ (s) + 2 NH₄Cl (aq)
Left Side
Mn =1
Cl = 2
N = 2
H = 8
C = 1
O = 3
Right Side
Mn =1
Cl = 2
N = 2
H = 8
C = 1
O = 3
Which statement best describes information that would be associated with science rather than with pseudoscience?
Answer:
The information has been changed to support a claim
Explanation:
what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
How to treat stream water for drinking
Answer:
Explanation:
Boiling.
Use water filter
Use Ultraviolet Light.
Use chlorine drops
I would recomade boiling as the main
because its the easiest and cheapest Or water filter if you have one
Reactive oxygen species (ROS) are unstable, or reactive, compounds that result from the partial reduction of oxygen. ROS can cause damage to molecules, including membrane lipids and nucleic acids, and may be associated with some diseases. Which of these compounds are reactive oxygen species? Choose all that apply.
a. OH
b. OH-
c. O2-
d. H2O
e. H2O2
f. H-
What type of bond will be formed for atoms that have a +1 or -1 charge?
Indicate whether the following represents a Chemical or Physical change: Milk sours
Answer:
Chemical Change
Explanation:
Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of the visible spectrum. Suppose a particular cone cell absorbs light with a wavelength of 434.nm. Calculate the frequency of this light. Round your answer to 3 significant digits.
Answer:
6.91 × 10¹⁴ s⁻¹
Explanation:
Step 1: Given data
Wavelength of the radiation absorbed by the cone (λ): 434 nm
Step 2: Convert the wavelength to meters
We will use the relationship 1 m = 10⁹ nm.
[tex]434nm \times \frac{1m}{10^{9}nm } =4.34 \times 10^{-7} m[/tex]
Step 3: Calculate the frequency (ν) of the radiation
We will use the following expression.
[tex]c = \lambda \times \nu[/tex]
where,
c is the speed of light (3.00 × 10⁸ m/s)
[tex]c = \lambda \times \nu\\\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}m/s }{4.34 \times 10^{-7}m }= 6.91 \times 10^{14} s^{-1}[/tex]
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO 3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH° rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.
Answer:
THE STANDARD HEAT OF NEUTRALIZATION OF THE BASE SODIUM HYDROXIDE BY THE ACID HYDROGEN TRIOXONITRATE V ACID IS -56 kJ / mol.
Explanation:
Volume of 0.3 M NaOh = 100 mL
Volume of 0.3 M HNO3 = 100 mL
Initail temp of NaOH and HNO3 = 35 °C = 35 + 273 K = 308 K
Final temp. of mixture = 37 °C = 37 + 273 K = 310 K
We can make the following assumptions form the question given:
1. specific heat of the reaction mixture is the same as the specific heat of water = 4.2 J/g K
2. the toal mass of the reaction mixture is 200 mL = 200 g since no heat is lost to the calorimeter or surrounding.
3. initail temperature of the reaction mixture is equal to the average temperature of the two reactant solutions
= ( 308 + 308 /2) = 308 K
4. Rise in temeperature for the reaction = 310 -308 K = 2 K
Then the total heat evolved during the reaction = mass * specifc heat capacity * temperature change
Heat = 200 g * 4.2 J/g K * 2 K
Heat = 1680 J
EQUATION FOR THE REACTION
HNO3 + NaOH -------> NaNO3 + H20
From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to prouce mole of water.
100 mL of 0.5 M HNO3 contains 100 * 0.3 /1000 = 0.03 mole of acid
This result is same for the base NaOH = 0.03 mole of base
So therefore,
0.03 mole of acid will react with 0.03 mole of base to produce 0.03 mole of water to evolved 1680 J of heat energy.
The production of 1 mole of water will evolve 1680 / 0.03 J of heat
= 56 000 J or 56 kJ of heat energy per mole of water.
So therefore, 1the standard heat of neutralization of sodium hydroxide by trioxoxnitrate V acid is -56 kJ/mol.
What are the relations between Electrochemistry and Cancer?
Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness
just like it does with chemical phenomena
=)
A stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Answer:
0.067M Fe(NO3)3
Explanation:
A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.
The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.
The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3
That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:
0.200M / 3 =
0.067M Fe(NO3)3The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.
Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]
[tex]k_b[/tex] = boiling point constant = ?
m = molality
[tex]w_2[/tex] = mass of solute (urea) = 55.4 g
[tex]w_1[/tex] = mass of solvent X = 500 g
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]
[tex]k_b=2.4^0C/m[/tex]
Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
A salt solution was found to contain 1.50 g of salt dissolved in 50 mL of water. On evaporation, the recovered salt weighed 1.47 g. What percent of salt was recovered?
A) 20.4%
B) 107%
C) 98%
D) 20.0%
Answer:
C = 98%
Explanation:
Hello,
To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.
Mass of salt recovered = 1.47g
Initial mass of salt = 1.50g
Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100
Percentage of salt recovered = (1.47 / 1.50) × 100
Percentage of salt recovered = 0.98 × 100
Percentage of salt recovered = 98%
The percentage of salt recovered is equal to 98%
ilicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). Masses and natural abundances for two isotopes are listed here. Isotope Mass (amu) Abundance (%) Si-28 27.9769 92.2 Si-29 28.9765 4.67 Si-30 ? ? Part A Find the natural abundance of Si-30 . Express your answer using two significant figures.
Answer:
Part A
The natural abundance of Si-30 = 3.1
Part B
The mass of Si-30 = 29.9551 amu
Explanation:
Part A
The sum of the natural abundances = 100
Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.
Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%
Si-30's mass and natural abundance are unknown.
Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.
Part B
The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.
Molar mass of Silicon normally = 28.0855 amu
Let the mass of Si-30 be m
28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)
28.0855 = 27.14790435 + 0.0313m
0.0313m = 28.0855 - 27.14790435 = 0.93759565
m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu
Hope this Helps!!
Part A: The natural abundance of Si-30 = 3.1
Lets solve the question:
The sum of the natural abundances = 100 Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%. Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67% Si-30's mass and natural abundance are unknown.Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.
Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.
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The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.
Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]
Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]
The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
[tex]K_a[/tex] for [tex]CH_3COOH[/tex] :
[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]
[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]
[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
