The correct statement about the description of ion whose symbol is Ca2+ among the choices above are:
It is a cationIt is a divalent metalThe options b and d are correct.
How calcium ion is a cation and divalent metalIt follows that elements whose charge on them are positive are called cations. In order words, in the periodic table of science of chemistry, the positively charged ions of elements are known as cations. Calcium has a positive of 2 on as it ion and this makes it a cation.
In addition to the above, calcium ion helps in signal transduction pathways of cells of living organisms. This goes to say that they also help in the body's normal functioning of the human body.
However, on a second note, calcium in the periodic table belongs to group 2, has a valence electrons of 2 and a metallic element; thereby making it a divalent metal.
In conclusion, we can now confirm from all said above that calcium above is a cation.
Complete question:
Check all of the following statements that are true about the ion whose symbol Ca2+
a. It is an anion
b. It is a cation
c. Its atoms is connected by tripple bonds
d. It is a divalent metal
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Determine whether each combination will produce aprecipitate. If a precipitate forms, enter its chemicalormula. If no precipitate forms, enter NONEHg2(NO3)2(aq) + BaCl₂(aq)H₂SO4(aq) + NaOH(aq)AgCH3CO₂(aq) + Cs₂SO4(aq)NH4Br(aq) + K3PO4(aq)
First, we have to complete the reactions with the respective products:
[tex][/tex]important characteristics of homologous series
EXPLANATION:
Homologous series can be defined as a sequence of compounds with the same functional group and similar chemical properties in which the members of the series differ by -CH2. The homologous series the chemical properties of the element.
Below are the characteristics of the homologous series
1. The members of the homologous series have the same functional group
2. The members of the homologous series have a general method of preparation
3. The members of the homologous series differ by -CH2
4. The members of the homologous series have the same chemical properties.
A sample of nitrogen gas, occupies 50.0 mL at 27 °C and 500. torr. What pressure will it have if cooled to –73 °C while the volume remains constant?
Answer:
333.33 torr
Explanation:
Initial Pressure = 500 torr
Let Final Pressure be 'P'
Initial Temperature = 27 °C = 300 K
Final Temperature = -73 °C = 200 K
Initial Volume = Final Volume
From Combined Gas Law,
(Initial Pressure × Initial Volume) / Initial Temperature = (Final Pressure × Final Volume) / Final Temperature
So, P = (500 torr × 200 K) / 300 K = 333.33 torr
In a 250 mL sealed container at 150℃, 0.50 mol of both iodine gas I2(g) and bromine gas, Br2(g) are mixed and allowed to react until they form an equilibrium with iodine monobromide gas IBr(g). The equilibrium constant for this reaction is 1.2 x 102. What are the equilibrium concentrations for iodine gas and bromine gas? I2(g) + Br2(g) ↔ 2IBr(g)
answer and explanation
now that we have the x value we can find the equilbrium concentratoins of tehegases
for Iodine
x =0.456
= 0.5-x
= 0.5-0.456
= 0.035mols
concentration = 0.035mols/ 0.250L
= 0.14M
for Bromine
x =0.456
= 0.5-x
= 0.5-0.456
= 0.035mols
concentration = 0.035mols/ 0.250L
= 0.14M
the concentration of the two gases is 0.14M
Draw the Lewis structure for chloroform, CHCl3
Answer:
Explanation:
To write the Lewis structure, it is important to look for the electrons of each atom in the Periodic Table of Elements.
C has 4 valence atoms, H has 1 valence atom and Cl has 7 valence atoms.
We write the C atom in the center. Then we have to complete the Octet in each atom, except in Hydrogen, because it can only has 2 electrons bonded.
The half-life of cobalt-60 is 10.47 min. How many grams of cobalt-60 remain after 104.7 min if you start with 1024g?
1.00147g of of cobalt-60 remain after 104.7 min if you start with 1024g.
According to nuclear decay [tex]N_{t} = N_{0 } e^[/tex]^-λt
for half-life ----[tex]t_{\frac{1}{2} }[/tex] = 0.693/λ
half-life of cobalt [tex]t_{\frac{1}{2} }[/tex] = 10.47
initial amount [tex]N_{0}[/tex] = 1024g.
t = 104.7 min
final product [tex]N_{t}[/tex] =?
[tex]t_{\frac{1}{2} }[/tex] = 0.69/λ ( according to half-life)
10.47 = 0.693/λ
λ = 0.693/10.47 [tex]min^{-1}[/tex]
[tex]N_{t} = N_{0 } e^[/tex]^-λt
[tex]N_{t}[/tex] = 1024[tex]\times e^{\frac{-0.693}{10.47} } \times 104.7 min[/tex]
[tex]N_{t}[/tex] = 1024 g[tex]\times e^{-6.93}[/tex]
[tex]N_{t}[/tex] = 1.00147g
What is half-life?Half-life is the time it takes for a quantity to decrease to half its original value. This term is often used in nuclear physics to describe how quickly unstable atoms decay radioactively or how long stable atoms last. The term is also used more generally to describe any type of exponential decay.
The half-life of a radioactive isotope is the time required for half of the radioactive isotope to decay. The half-life of a given radioactive isotope is constant. it is unaffected by conditions and is independent of the initial amount of that isotope.
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The designations for the first four electron energy sublevels with the maximum number of electrons that can be accommodated in each sublevel ares:2, p:6, d:10, and f:14s:2, p:8, d:10, and f:14s:1, p:2, d:3, and f:4s:2, p:8, d:18, and f:32
So,
The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max. And the 4 sublevel has 7 orbitals, so can contain 14 electrons max.
Therefore, the correct answer is:
s:2, p:6, d:10, and f:14
Initially a sample of nitrogen gas is found in a pressure of 4.62 atm under the temperature of 27.0°C
The change in the temperature of the gas, given that the volume is held constant is 63 °C
How to determine the change in temperatureLet's begin by obtaining the new temperature of the gas. This can be obtained as follow:
Initial pressure (P₁) = 4.62 atm Initial temperature (T₁) = 27 °C = 27 + 273 = 300 KNew pressure (P₂) = 5.59 atm New temperature (T₂) =?P₁ / T₁ = P₂ / T₂
4.62 / 300 = 5.59 / T₂
Cross multiply
4.62 × T₂ = 300 × 5.59
Divide both side by 4.62
T₂ = (300 × 5.59) / 4.62
T₂ = 363 K
Subtract 273 to obtain answer in °C
T₂ = 363 – 273 K
T₂ = 90 °C
Finally, we shall determine the change in the temperature. This is illustrated below:
Initial temperature (T₁) = 27 °CNewl temperature (T₂) = 90 °CChange in temperature (ΔT) =?ΔT = T₂ – T₁
ΔT = 90 – 27
ΔT = 63 °C
The, the change in the temperature is 63 °C
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Complete question:
See attactched photo
I need help! Thank u so so much in advance
Electron capture occurs when one of the electrons in the atom is captured by the atom's nucleus. The captured electron combines with one proton to form a neutron.
We have to find the nuclide that is produced when the aluminum-28 undergoes an electron capture.
If we look for the atomic number of Al in the periodic table we will see that it is 13. That means that it has 13 protons. Then we are given the atomic mass of aluminum, it is 28. That means that if we add the number of protons and the number of neutrons we get 28.
atomic mass = n° of protons + n° of neutrons = 13 + 15 = 28
atomic number = n° of protons = 13
The nucleus of the aluminum atom will capture one electron that will combine with a proton of the nucleus to form a neutron. So the atomic number will change because it will have 1 proton less. But the atomic mass won't change because we are losing one proton but winning a neutron, so it remains the same. After the electron capture, the new nuclide will have:
atomic mass nuclide = n° of protons + n° of neutrons = 12 + 16 = 28
atomic number nuclide = n° of protons = 12
If we look for the element that has an atomic number of 12 in the periodic table we will see that it is magnesium.
Finally we can express this like a reaction:
28 0 28
Al + e --------> Mg
13 -1 12
Answer:
28
Mg
12
If 3.8 moles of P4 are used, how many moles of O2 are needed?
1) Write the chemical equation.
[tex]P_4+O_2\rightarrow P_4O_{10}[/tex]2) List the elements in the reactants and in the products.
Reactants.
P: 4
O: 2
Products
P: 4
O: 10
Balance O.
[tex]P_4+5O_2\rightarrow P_4O_{10}[/tex]3) Moles of O2 needed.
The molar ratio between P4 and O2 is 1 mol P4: 5 mol O2.
[tex]mol\text{ }O_2=3.8\text{ }mol\text{ }P4*\frac{5\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }P_4}=19\text{ }mol\text{ }O_2[/tex]If 3.8 mol P4 are used, 19 mol O2 are needed.
.
calculate the number of litres of each gas givin the following number of moles at STPhow many litres of gas is in 0.001 mol H2S
STP stands for Standard Temperature and Pressure. Nowadays the STP is temperature of 0°C (273.15K) and pressure of 1 bar.
Assuming the gas behaves as an ideal gas, in these conditions the volume of each mol of gas is approximately 22.7 L.
Using rule of three, we can calculate the volume of 0.001 mol of H₂S:
Volume --- Number of moles
x --- 0.001 mol
22.7L --- 1 mol
[tex]\begin{gathered} \frac{x}{22.7L}=\frac{0.001}{1} \\ x=22.7\cdot0.001L \\ x=0.0227L \end{gathered}[/tex]So, the volume is approximately 0.0227 L.
How many ions in 800.0mL of 0.10M sucrose
M = mol/L
To solve for the number of atoms in a certain amount of moles, we must multiply the number of moles by Avogadro's number ([tex]N_A[/tex]): [tex]6.02*10^{23}[/tex], and multiply the product by the number of atoms in the compound.
Solving the QuestionWe're given:
Sucrose = [tex]C_{12}H_{22}O_{11}[/tex]800.0 mL0.10 MFirst, we must find the number of moles. We can do this by first converting 800.0 mL to L and multiplying it by 0.10 M.
800.0 mL ÷ 1000 = 0.8 L
⇒ Now, multiply 0.8 L by 0.10 M:
0.8 L × 0.10 M = 0.08 mol
Now, we must find the number of molecules in this sample. Multiply 0.08 mol by Avogadro's number:
[tex]0.08*6.02*10^{23}\\=0.4816*10^{23}\\=4.816*10^{22}[/tex]
Finally, to find the number of atoms in the sample, multiply the number of molecules by the number of atoms in the compound.
We can begin by finding the number of atoms in sucrose:
[tex]C_{12}H_{22}O_{11}[/tex]
12 + 22 + 11 = 45 atoms
⇒ Multiply 45 atoms by [tex]4.816*10^{22}[/tex] mol:
[tex]45*4.816*10^{22}\\=216.72*10^{22}\\=2.1672*10^{24}[/tex]
Convert to the correct number of significant figures (2):
[tex]=2.17*10^{24}[/tex]
Answer[tex]2.17*10^{24}[/tex] atoms
Please help me to find yield of Alcl3 in gram
Theoretical yield is calculated based on the stoichiometry of the chemical equation. First, we have to find the number of moles of each reactant using the molar mass of each of them. You can find the molar mass using the periodic table: the molar mass of aluminum (Al) is 26.98 g/mol and the molar mass of chlorine gas (Cl2) is 70.8 g/mol:
[tex]34.0\text{ g Al}\cdot\frac{1\text{ mol Al}}{26.98\text{ g Al}}=1.26\text{ moles Al.}[/tex][tex]39.0gCl_2\cdot\frac{1\text{ mol }Cl_2}{70.8\text{ g }Cl_2}=0.551\text{ moles }Cl_2.[/tex]The next step is to find how many moles of AlCl3 are produced by each reactant. You can see that in the chemical equation, 2 moles of Al produces 2 moles of AlCl3:
[tex]1.26\text{ moles Al}\cdot\frac{2molesAlCl_3}{2\text{ moles Al}}=1.26molesAlCl_3.[/tex]And 3 moles of Cl2 produces 2 moles of AlCl3:
[tex]0.551molesCl_2\cdot\frac{2molesAlCl_3}{3molesCl_2}=0.367molesAlCl_3.[/tex]You can realize that the maximum, amount of moles that are produced of AlCl3 is 0.367 moles because Cl2 would be the limiting reactant and there is an excess of Al. For this value, we have to find its mass using the molar mass of AlCl3 which is 133.2 g/mol:
[tex]\text{0}.367molesAlCl_3\cdot\frac{133.2gAlCl_3}{1molAlCl_3}=48.88gAlCl_3.[/tex]The answer would be that the theoretical yield of AlCl3 is 48.88 g.
How many milliliters of 0.100M NaOH are required to neutralize the following solution 5.00mL of.0500M HCI
1) Set the chemical equation
[tex]\text{NaOH + HCl = NaCl + H}_2O[/tex]2) Find out how many moles of HCl react in the solution.
[tex]\text{moles of HCl = }\frac{0.0500\text{ M}}{\square}\cdot\frac{5.00\text{ mL}}{\square}\cdot\frac{1\text{ L}}{1000\text{ mL}}=\text{ 0.00025 moles of HCl}[/tex]3) Find out how many moles of NaOH reacted with HCl
[tex]\text{moles of NaOH = }\frac{0.00025\text{ moles of HCl}}{\square}\cdot\frac{1\text{ mole of NaOH}}{1\text{ mole of HCl}}=\text{ 0.00025 moles of NaOH}[/tex]4) Find out the milliliters of NaOH
[tex]mL\text{ of NaOH = }\frac{0.00025\text{ moles of NaOH}}{0.100\text{ M}}\cdot\frac{1000\text{ mL}}{1\text{ L}}\text{= 2.50 mL}[/tex]1. Find the elements sodium, oxygen, and phosphorus on the periodic table. Describe the monoatomic ions each would form. Then, given what you know about chemical bonding, describe and give the formulas for the types of ionic compounds they would produce with each other. Use the periodic table in the Chemistry: Problems and Solutions book to help with your answers.2. Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.KIFe2O3(NH4)3PO4
Answer for 2:
- KI: potassium iodide
- Fe2O3: diiron trioxide
- (NH4)3PO4: ammonium phosphate
Explanation for 2:
• KI,:
The compound KI is called potassium iodide, because it is a hydracid salt that has potassium in the positive parte (that's why it keeps the name of the metal). While the negative parte of the salt comes from the non-metal iodine.
• Fe2O3,:
The compound Fe2O3 is called
, because the molecule of this oxide has 3 oxygen atoms and 2 iron atoms.
• (NH4)3PO4:
The (NH4)3PO4 is a salt called ammonium phosphate, because it is formed from phosphoric acid (that gives the negative part of the salt), and ammonia (that gives the positive part of the salt). The negative part of salt ends in "...ate" because it comes from the acid that ends with the suffix "...ic".
If the volume is 10 and the mass of water is 9.99 what is the density
Answer:
0.999
Explanation: divide by 10
Calculate the frequency of light that has a wavelength of 16.23 * 10^-9 m.
The frequency of the light that has a wavelength of 16.23 × 10⁻⁹ meter would be 1.84 × 10¹⁶ Hz , as the frequency and the wavelength are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem we have to calculate the frequency of the light if the wavelength of light is 16.23 × 10⁻⁹ meter.
The frequency of the light = 3 × 10⁸ / 16.23 × 10⁻⁹
= 1.84 × 10¹⁶ Hz
Thus, the frequency of the light would be 1.84 × 10¹⁶ Hz.
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a gas has a volume of 103.4 L at 27.5 atm. what will it’s volume be if the pressure is changed to 69.3 atm
In this question, we have a gas presenting volume and pressure, and the gas law that we can use to calculate it is Boyle's gas Law, which perfectly correlates both pressure and volume, the formula is:
P1V1 = P2V2
We have:
P1 = 27.5 atm
V1 = 103.4 L
P2 = 69.3 atm
V2 = ?
Now we add these values into the formula:
27.5 * 103.4 = 69.3 * V2
V2 = 41.0 L
How many gram of iron is how many moles in 5237 & how many in 4180
To get the number of moles, you need to use this equation:
n = m/M
where n is the number of moles, m is the mass and M is the molar mass of Iron.
Given:
m = 5227 g
M = 55.8 g/mol (from the periodic table)
Therefore
n = 5227 g/55.8 g/mol
n = 93.7
The first option is closer to 93.7 so the answer is the first option.
LT.3c - REDOX - QUESTION 4The following is a redox reaction. Label all oxidation states anddetermine which element is being oxidized and which element isbeing reduced. Briefly explain your answers in words.Fe2O3 + СО -> Fe + CO2
Answer: The oxidation states can be labelled as:
- In Fe2O3: Fe = +3, O = -2
- In CO: C = +2, O = -2
- In Fe: Fe = 0
- In CO2: C = +4, O = -2
Fe is being reduced (changes from +3 to 0) and C is being oxidized (changes from +2 to +4).
Explanation:
The question requires us to label the oxidation states of each element in the following chemical equation, and then determine which one is being oxidized and which one is being reduced.
[tex]Fe_2O_3+CO\rightarrow Fe+CO_2[/tex]To solve this problem, we need to keep in mind a few points:
- oxidation occurs when its oxidation number increases (the element "loses" an electron);
- reduction occurs when its oxidation number decreases (the element "gains" an additional electron);
- the oxidation number of some elements is usually the same; a good example is oxygen (O): in most cases, O presentes oxidation number equals to -2.
Next, let's analyze the oxidation numbers of the elements in the reaction:
- Fe2O3: since O has oxidation number = -2, in the compound Fe2O3 the total charge brought by O is (-2) * 3 = -6, thus 3 atoms of Fe must have charge +6. We can say Fe has oxidation number = +3 in Fe2O3 (because (+3) * 2 + (-2) * 3 = 0).
- CO: O has oxidation number = - 2, thus C must present oxidation number = +2 in CO.
- Fe: Elementar Fe presents oxidation number = 0;
- CO2: Since O has oxidation number = -2, C must present oxidation number = +4 (because (+4) * 1 + (*2( * 2 = 0.
Thus, the oxidation states can be labelled as:
- In Fe2O3: Fe = +3, O = -2
- In CO: C = +2, O = -2
- In Fe: Fe = 0
- In CO2: C = +4, O = -2
Analyzing the oxidation states in the reactants (left side) and products (right side), we can see that Fe goes from +3 to 0, thus it is being reduced, while C goes from +2 to +4, thus it is being oxidized.
The 946 mL of Gatorade bottle contains FD&C Blue I at concentration of 0.100mg/L. Molar mass of Blue I is 792.84 g/mol.
a) Calculate the molarity of Blue I in Gatorade.
b) How many molecules of Blue I would you consume by drinking the entire bottle of Gatorade.
The molarity of Blue I in Gatorade is 0.00010 mol L⁻ molecules of Blue I would you consume by drinking the entire bottle of Gatorade is 4.77 mol
Molarity is the amount of substance in certain volume of solution and also moles of solute per liter of solution
Here given data is
Volume = 946 mL
Molar mass = 792.84 g/mol
Concentration = 0.100mg/L
We have to find molarity and molecules of Blue I would you consume by drinking the entire bottle of Gatorade = ?
Molarity = number of moles of solute/volume of solution
Molarity = 0.100mg/L/946 mL
Molarity =0.00010 mol L⁻
Now we have to find molecules of Blue I = ?
Molecules of Blue I = 792.84 g/mol × avagadro number
Avagadro number = 6.022×10²³
Molecules of Blue I = 792.84 g/mol×6.022×10²³
Molecules of Blue I = 4.77 mol
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A mixture of nitrogen and hydrogen gas is compressed from a volume of 90.0L to a volume of 89.0L, while the pressure is held constant at 44.0atm. Calculate the work done on the gas mixture. Round your answer to 2 significant digits, and be sure it has the correct sign (positive or negative).
The PV work formula is:
w = -Pexternal * ΔV
w = -44.0 atm * (89.0 - 90.0)
w = -44.0 atm * -1 L
w = 44.0 L atm
For each L atm, we will have 101.325 Joules, therefore
44.0 L atm * 101.325 J
w = 4458.3 J or 4.4 kJ
How many moles are there in 2.50 grams of ethanol, C2H6O ?
What is the name for MgBr2?Magnesium bromideMagnesium bromineMagnesium dibromideDimagnesium bromide
The name for MgBr2 is Magnesium bromide
Consider this reaction:3Ca(s)+2H3PO4(aq)=Ca3(PO4)2(s)+3H2(g)How many grams of calcium are required to produce 50.500 g of calcium phosphate (MW = 310.19g/mol) assuming an excess of Phosphoric acid?
Answer
19.6 grams
Explanation
Given:
Mass of calcium phosphate produced = 50.500 g
Equation: 3Ca(s) + 2H3PO4(aq) ---> Ca3(PO4)2(s) + 3H2(g)
What to find:
The grams of calcium required to produce 50.500 g of calcium phosphate.
Step-by-step solution:
From the equation, 3 mol Ca produce 1 mol Ca3(PO4)2
1 mole Ca3(PO4)2 = 310.19 grams
1 mole Ca = 40.078 grams
This means, (3 x 40.078 g) = 120.234 g Ca produce 310.19 g Ca3(PO4)2
So x grams Ca will be required to produce 50.500 grams Ca3(PO4)2
x grams Ca will be equal
[tex]\frac{50.500\text{ }g}{310.19\text{ }g}\times120.234g\text{ }Ca=19.6\text{ }grams\text{ }Ca[/tex]The grams of calcium required to produce 50.500 g of calcium phosphate = 19.6 grams
Calculate the number of molecules in 2.37 g of c3H8 (molar mass 44.11 g/mol)
Step 1
Avogadro's number states that:
1 mole of C3H8 = 44.11 g of C3H8 = 6.02x10^23 formula units (molecules in this case)
---------
Step 2
Information provided:
2.37 g C3H8
The molar mass = 44.11 g/mol
---------
Step 3
Procedure:
44.11 g C3H8 ---------- 6.02x10^23 molecules of C3H8
2.37 g C3H8 ---------- X
X = 2.37 g C3H8 x 6.02x10^23 molecules of C3H8/44.11 g C3H8
X = 3.23x10^22 molecules of C3H8
Answer: Please, revise your options because I cannot find any mistake in my procedure. (Ask your teacher, please)
I found 3.23x10^22 molecules of C3H8 but there is no option alike.
Answer the following questions assuming you have a 1.3M (molarity) solution ofHNO3 (5 points each, 15 points total)A. How many moles of HNO3 would you have in 0.6 liters of the above solution?B. How many grams of HNO3 would be present in 0.6 liters of the above solution?C. If you began with 0.6 liters of solution, and diluted the solution to a concentration of 0.45M, whatwould the final volume of the solution be?
A) 0.78 moles
B) 49.15grams
C) 1.73L
Explanations:The formula for calculating the molarity of a solution is expressed as;
[tex]\begin{gathered} molarity=\frac{moles}{volume} \\ moles=molarity\times volume \end{gathered}[/tex]Given the following parameters
molarity of HNO3 = 1.3M
volume of HNO3 = 0.6L
A) moles of HNO3 = 1.3 * 0.6
moles of HNO3 = 0.78moles
B) Mass of HNO3 = moles * molar mass
Mass of HNO3 = 0.78 * 63.01
Mass of HNO3 = 49.15grams
There are 49.15grams of HNO3 present in 0.6L of the solution
C) According to dilution formula
[tex]\begin{gathered} C_1V_1=C_2V_2 \\ V_2=\frac{C_1V_1}{C_2} \\ V_2=\frac{1.3\times0.6}{0.45} \\ V_2=\frac{0.78}{0.45} \\ V_2=1.73L \end{gathered}[/tex]Therefore the final volume of the solution will be 1.73L
When copper is heated with an excess of sulfur, copper(I) sulfide is formed.
In a given experiment, 1.53 g of copper was heated with excess sulfur to yield 1.76 g of copper(I) sulfide.
What is the percent yield?
If 1.53 g of copper is heated with excess sulfur to yield 1.76 g of copper (I) sulfide, the percent yield would be 75.43%.
Percent yieldThe percent yield of a reaction is the actual yield relative to the theoretical yield of the same reaction.
The percent yield of a reaction is mathematically expressed as:
Percent yield = actual yield/theoretical yield x 100%.
Let us now consider the reaction in question. The equation of the reaction is given as:
[tex]Cu + S --- > CuS[/tex]
The mole ratio of the copper that reacts to the copper sulfide that is formed is 1:1.
Recall that: mole = mass/molar mass
The molar weight of copper is 63.55 g/mol
Mole of 1.53 g copper = 1.53/63.55
= 0.024 moles
The equivalent mole of copper sulfide produced will also be 0.024 moles.
The molar mass of CuS is 95.61 g/mol
Mass of 0.024 moles CuS = 0.024 x 96.61
= 2.32 g
2.32 g is the theoretical yield of the reaction. But 1.76 g of copper sulfide is the actual yield.
Percent yield = 1.75/2.32 x 100%
= 75.43%
Thus, the percent yield of the reaction, if 1.53 g of copper is heated with excess sulfur to produce 1.76 g of copper (I) sulfide, is 75.43%.
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Ozone molecules in the upper atmosphere absorb radiation. If the radiation has a wavelength between 240 nm and 310 nm, the ozone molecules will decompose into oxygen molecules and oxygen atoms. The oxygen atoms then recombine with the oxygen molecules to make more ozone, releasing heat. This converts light energy into heat energy and insulates the Earth. ***** O3(g) → O2(g) + O(g) -> light energy absorbed heat energy released O2(g) + O(g) 03(g) a) What kind of electromagnetic radiation has a wavelength between 240 and 310 nm? b) Which wavelength represents the minimum amount of energy required for this reaction to proceed: 240 nm or 310 nm? c) Calculate the minimum amount of light energy that must be absorbed to convert 1 mole of ozone into oxygen molecules and atoms. Report your answer in kJ/mol. [4 Marks]
a)ultraviolet rays
Explanation:
ultraviolet rays are electromagnetic waves that are found in the range of 240 and 310 somewhere in between the electromagnetic spectrum
b)310nm causes minimum energy because the ultraviolet rays have reacted with oxygen to produce zone
c) 310nm to
=3.1×10^-7m once converted find the frequency using the converted wavelength as your wavelength AND once that has been done find your
E=hv=6.626×10^-34×9.68×10^14=6.414×10^-10J convert J to KJ/MOL will be 386.3KJ/MOL.
A patient receives 100 mL of 25% (m/v) mannitol solution every hour.How many grams of mannitol are given in 1 hour?Express your answer using two significant figures.
Explanation:
A patient receives 100 mL of a 25 % m/v mannitol solution every hour. We have to find the mass in grams of mannitol that the pacient is receiving in 1 hour. We can use the formula of the mass/volume percent to find the answer to our problem.
% m/v = mass of solute in g/(volume of solution in mL) * 100 %
We know the concentration of the solution and we have the volume of solution in mL, so we can solve the formula for the mass of solute and get the answer to the problem.
25 % m/v = mass of solute in g/100 mL * 100 %
mass of solute in g = 25 * 100/100 g
mass of solute in g = 25 g
Answer: In 1 hour the patient receives 25 g of mannitol.