Explanation:
3Ca(s) + 2AlCl3(aq) -> 3CaCl2(aq) + 2Al(s)
According to the question, Ca is the limiting reactant.
Therefore, we equate Ca to Aluminium which is the product whose mass we want to find
Molar mass of Ca- 40g/mol
". ". of Al- 27g/mol
3Ca --> 2Al
3×40 --> 2×27
9.2 --> x
x = 9.2×2×27= 496.8÷120=4.14
The steps of preparing 2M solution of NaOH
Answer:
And 500 mL of 2M NaOH solution contains 80.02g NaOH=40.0 g of NaOH .
Explanation:
So, If we mix 40.0 g of NaOH with enough distilled water to make 500 mL, we will get a 2.00 M NaOH solution.
I hope this answers help you! :>
1. What do you understand about equivalent weight for lime requirement?
2. What is the relationship between normality and equivalence?
Answer:
Explanation:
What do you understand about equivalent weight for lime requirement?
If field data are not available, lime requirements can be determined with the solution is added to a given weight or volume of soil and the pH is measured. Each milligram per liter of sodium bicarbonate is equal to 0.56 mg/L of alkalinity. A winter wheat crop grown on a light sand soil in a low rainfall area will mean an
The change in entropy of 20.0 moles of ideal monatomic constant volume gas as it changes from an initial temperature 300 K to its final temperature is 200 J/K. What is the final temperature of the gas
Answer:
668.9K is the final temperature
Explanation:
The change in entropy, ΔS, of an ideal monoatomic gas is obtained using the equation:
ΔS = n*Cv*ln (T2/T1)
Where ΔS is change in entropy = 200J/K
n are moles = 20.0mol
Cv is 3/2R for an ideal monoatomic gas (3/2*8.314J/molK)
T2 is final Temperature and T1 initial temperature = 300K
Replacing:
ΔS = n*Cv*ln (T2/T1)
200J/K = 20.0mol*3/2 *8.314J/molK*ln (T2/300K)
0.80186 = ln (T2/300K)
2.23 = T2 / 300K
668.9K is the final temperatureHELP HELP HELP
NH3 + NO + N2 + H2O
5. Given 8.25 x 1025molecules of ammonia, determine the number of grams of
nitrogen produced.
Answer:
4NH3+6NO+5N2+6H20
Explanation:
Question 1
3 pts
TI
Which part(s) of the following ground state electron configuration holds the valence
electrons?
At
o
1s22s22p63523p4
ОЗр
03s
O2s
O 35 and 3p
Answer:
3s and 3p
Explanation:
From the question given above, the following data were obtained:
Electronic configuration =>
1s² 2s²2p⁶ 3s²3p⁴
Location of valence electron =?
From the electronic configuration given above, we can see clearly that the atom has three (3) shells.
Valence electron(s) are located at the outer most shell of an atom.
The outer most shell of the atom above is 3s and 3p.
Therefore, 3s and 3p will contain the valence electron(s)
How many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base, to produce 1.00 L of a buffer that is
Answer:
79.1g of weak base must be combined with 56.0g of conjugate acid
Explanation:
that is 1.00 M in the weak base?
The weak base is C5H5N with a pKa of 5.17 (Ka=6.7×10-6)and a desire pH of 5.63
The equilibrium of the weak base is with the bromide salt of the conjugate acid is:
C5H5N(aq) + H2O(l) + Br- ⇄ C5H5NHBr(aq) + OH-(aq)
Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.7x10⁻⁶
Kb = 1.49x10⁻⁹ is defined as:
Kb = 1.49x10⁻⁹ = [C5H5NHBr] [OH-] / [C5H5N]
Where [OH-] = 10^-(14- pH) = 10^-(14- 5.63) = 4.255x10⁻⁹M
[C5H5N] = 1.00M
Replacing:
1.49x10⁻⁹ = [C5H5NHBr] [OH-] / [C5H5N]
1.49x10⁻⁹ = [C5H5NHBr] [4.255x10⁻⁹M] / [1.00M]
[C5H5NHBr] = 0.35M
In 1L the moles of C5H5NHBr are 0.35 moles
Mass C5H5NHBr: 160.0118g/mol
0.35 moles * (160.0118g / mol) =
56.0g of C5H5NHBr are necessaries
The mass of C5H5N is -79.1g/mol-:
1.00moles * (79.1g/mol) =
79.1g of C5H5N are necessaries1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
2. (8pts) how much heat and kilojoules is needed to raise the temperature of 236 g of silver from 8.5°C to 34.9°C? (Cag = 0.24 J/g°C)
Answer:
1. 2.510kJ
2. Q = 1.5 kJ
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, we can proceed as follows:
1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:
[tex]1cal=4.184J\\\\1kJ=1000J[/tex]
Then, we perform the conversion as follows:
[tex]600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ[/tex]
2. Here, we use the general heat equation:
[tex]Q=mC(T_2-T_1)[/tex]
And we plug in the given mass, specific heat and initial and final temperature to obtain:
[tex]Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ[/tex]
Regards!
What is the mole ratio of O2 to NH3?
Write an equation for sodium chloride and sodium oxide
Answer:
Explanation:
The Chemical Equation for each one of these compounds would be the following
Sodium Chloride: NaCl
This compound contains 1 Sodium atom and 1 Chlorine atom
Sodium Oxide: Na₂O
This compound contains 2 Sodium atoms and 1 Oxygen atom.
1.00 x 10-7 moles to grams
If g of chlorine gas occupies a volume of mL at a particular temperature and pressure, what volume will g of chlorine gas occupy under the same conditions
The complete question is as follows: If 1.04 g of chlorine gas occupies a volume of 872 mL at a particular temperature and pressure, what volume will 2.08 g of chlorine gas occupy under the same conditions ?
Answer: A volume of 1744 L will 2.08 g of chlorine gas occupy under the same conditions.
Explanation:
Given: [tex]Mass_1[/tex] = 1.04 g, [tex]V_{1}[/tex] = 872 mL
[tex]Mass_2[/tex] = 2.08 g, [tex]V_{2}[/tex] = ?
As molar mass of chlorine is 35.5 g/mol.
Number of moles is the mass of a substance divided by its molar mass.
Hence, moles of chlorine present in 1.04 g chlorine gas is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{1.04 g}{35.5 g/mol}\\= 0.029 mol[/tex]
Also, moles of chlorine present in 2.08 g chlorine gas is calculated as follows.
[tex]Moles = \frac{2.08 g}{35.5 g/mol}\\= 0.058 mol[/tex]
Formula used to calculate the volume occupied by 2.08 g of chlorine gas is as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{872 L}{0.029 mol} = \frac{V_{2}}{0.058 mol}\\V_{2} = 1744 L[/tex]
Thus, we can conclude that a volume of 1744 L will 2.08 g of chlorine gas occupy under the same conditions.
determine the mass fraction of iron in its compounds:
Please!
Answer:
Fe(NO₃)₃ = 0.23; FeCl₃ = 0.34
Explanation:
Mass fraction is defined as the ratio between the mass of iron in 1 mol of the compound and the mass of the complete compound.
In Fe(NO₃)₃ there is just 1 atom of Fe that has a molar mass of 55.845g/mol. The molar mass of this compound is 241.86g/mol. Thus, mass fraction is:
55.845g/mol / 241.86g/mol =
Fe(NO₃)₃ = 0.23In FeCl₃ there is 1 atom of Fe that has a molar mass of 55.845g/mol. The molar mass of this compound is 162.2g/mol. Thus, mass fraction is:
55.845g/mol / 162.2g/mol =
FeCl₃ = 0.34Which of the following statements describes a double displacement reactions?
Answer:
D
Explanation:
hope this helped and sorry if it didn't but I wish you luck:D
In the lab, a 156 g sample of unknown compound KClOx was completely decomposed to produce 50.4 L of O2 (g) at STP and an unknown amount of KCl. What is the value of X
Answer:
2
Explanation:
Step 1: Write the generic decomposition reaction
KClOₓ ⇒ KCl + O₂
Step 2: Calculate the moles of O₂
At STP, 1 mole of an ideal gas occupies 22.4 L.
50.4 L × 1 mol/22.4 L = 2.25 mol
Step 3: Calculate the mass corresponding to 2.25 moles of O₂
The molar mass of O₂ is 32.00 g/mol.
2.25 mol × 32.00 g/mol = 72 g
Step 4: Determine the value of x
Every 156 g of KClOₓ there are 72 g of oxygen. The mass percent of O is:
%O: 72 g/156 g × 100% = 46.2%
In KClO, the mass percent of O is: 32.00 g/90.5 g × 100% = 35.4%
In KClO₂, the mass percent of O is: 64.00 g/138.5 g × 100% = 46.2%
Then, x = 2
Write the Ksp expression for the sparingly soluble compound cobalt(II) sulfide, CoS. If either the numerator or denominator is 1, please enter 1.
Answer:
[tex]Ksp=\frac{[Co^{2+}][S^{2-}]}{1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary to write the chemical equation for this reaction as shown below:
[tex]CoS(s)\rightleftharpoons Co^{2+}(aq)+S^{2-}(aq)[/tex]
Thus, since solids are not included in equilibrium expressions, we can set this one up as follows:
[tex]Ksp=\frac{[Co^{2+}][S^{2-}]}{1}[/tex]
Regards!
A balance was tested against a standard calibration mass with a certified value of 200.002 g and produced the following readings in grams: 195.587, 195.586, 195.588, 195.587, 195.589. The performance of this balance is best described as:
Answer:
Precise but not accurate.
Explanation:
We can tell the performance of the balance is precise, because the repeated measurements give values close to one another.
However, the performance of the balance is not accurate, as the mean value of the repeated measurements (195.587) is not close to the value considered as true (in this case the standard calibration mass with a certified value of 200.002 g).
Select three forms of non-renewable energy.
Biomass
Peat
Nuclear
Gas
Answer: Biomass, Nuclear and Gas are the three forms of non- renewable energy.
Explanation:
How many moles are in 150g of Li2O?
Answer:
moles= w/mm
moles=150/30
moles=5
Calculate the total energy (in kJ) absorbed when 50.5 g of ice at -15.0°C is converted into liquid water at 65.0 °C.
Answer:
The total energy absorbed is 32.171 kilojoules.
Explanation:
The total energy absorbed by the ice is the sum of the sensible heat of ice and water and the latent heat of fusion of the water, that is:
[tex]Q = m\cdot [c_{i}\cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})][/tex] (1)
Where:
[tex]m[/tex] - Mass of the ice, in kilograms.
[tex]c_{i}[/tex] - Specific heat of ice, in kilojoules per kilogram-degree Celsius.
[tex]c_{w}[/tex] - Specific heat of water, in kilojoules per kilogram-degree Celsius.
[tex]L_{f}[/tex] - Latent heat of fusion, in kilojoules per degree Celsius.
[tex]T_{1}[/tex] - Initial temperature of water, in degrees Celsius.
[tex]T_{2}[/tex] - Fusion point of water, in degrees Celsius.
[tex]T_{3}[/tex] - Final temperature of water, in degree Celsius.
[tex]Q[/tex] - Total energy absorbed, in kilojoules.
If we know that [tex]m = 50.5\times 10^{-3}\,kg[/tex], [tex]c_{i} = 2.090 \,\frac{kJ}{kg\cdot^{\circ}C}[/tex], [tex]c_{w} = 4.180\,\frac{J}{kg\cdot ^{\circ}C}[/tex], [tex]L_{f} = 334\,\frac{kJ}{kg}[/tex], [tex]T_{1} = -15\,^{\circ}C[/tex], [tex]T_{2} = 0\,^{\circ}C[/tex] and [tex]T_{3} = 65\,^{\circ}C[/tex], then the total energy absorbed is:
[tex]Q= (50.5\times 10^{-3}\,kg)\cdot \left[\left(2.090\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (15\,^{\circ}C) + 334\,\frac{kJ}{kg}+ \left(4.180\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (65\,^{\circ}C)\right][/tex][tex]Q = 32.171\,kJ[/tex]
The total energy absorbed is 32.171 kilojoules.
How many moles of RbNO2 are present in a sample with 3.4 x 1024 particles?
Answer:
5.6 mol
Explanation:
n = N/NA
n: number of moles
N: number of particles
NA: Avogadro's constant
n = 3.4×10²⁴/6.02×10²³ mol^-1 = 5.6 mol
The picture below shows a NASA image of the Oort cloud, a sphere of objects that are thought to surround Earth's solar system at a distance of up to one light-year from the Sun.
If the Oort cloud does exist, which of the following could explain why the objects in it may have formed this spherical shape?
A
They are held in orbits by the Sun's gravitational force.
B
They are held in Earth's solar system by Neptune's magnetic field.
C
They float freely in space because the Sun's gravitational force is too weak to hold them in orbit.
D
They are held in stationary positions around Earth's solar system by other stars in space.
The gravitational pull of the Sun keeps them in their orbits. The gravitational pull of the Sun and surrounding stars combined is most likely what gives the Oort cloud its spherical shape.
What does NASA's Oort Cloud mean?The Oort Cloud is a spherical layer of ice objects that is thought to be located between 2,000 and 100,000 astronomical units (AU) from our Sun, a star.
What makes it the Oort Cloud?The Oort cloud, so named after the Dutch astronomer Jan Oort who first confirmed its existence, is a collection of objects with a combined mass estimated to be 10-100 times that of Earth that are less than 100 km (60 miles) across and maybe trillions in number.
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Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for CaY2- is 5.0 x 1010 and the fraction of unprotonated EDTA (Y4-) is 0.35 at pH10. Calculate the concentration of free Ca2 in this solution.
Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
[tex]Ca^{2+[/tex] + [tex]y^{4-[/tex] ⇄ [tex]CaY^{2-[/tex]
Formation constant Kf
Kf = [tex]CaY^{2-[/tex] / ( [[tex]Ca^{2+[/tex]][[tex]y^{4-[/tex]] ) = 5.0 × 10¹⁰
Now,
[[tex]y^{4-[/tex]] = [tex]\alpha _4CH_4Y[/tex]; ∝₄ = 0.35
so the equilibrium is;
[tex]Ca^{2+[/tex] + [tex]H_4Y[/tex] ⇄ [tex]CaY^{2-[/tex] + 4H⁺
Given that; [tex]CH_4Y[/tex] = [tex]Ca^{2+[/tex] { 1 mol [tex]Ca^{2+[/tex] reacts with 1 mol [tex]H_4Y[/tex] }
so at equilibrium, [tex]CH_4Y[/tex] = [tex]Ca^{2+[/tex] = x
∴
[tex]Ca^{2+[/tex] + [tex]y^{4-[/tex] ⇄ [tex]CaY^{2-[/tex]
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = [tex]CaY^{2-[/tex] / ( [[tex]Ca^{2+[/tex]][[tex]y^{4-[/tex]] ) = [tex]CaY^{2-[/tex] / ( [[tex]Ca^{2+[/tex]][[tex]\alpha _4CH_4Y[/tex]] ) = 5.0 × 10¹⁰
⇒ [tex]CaY^{2-[/tex] / ( [[tex]Ca^{2+[/tex]][[tex]\alpha _4CH_4Y[/tex]] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
It is well-known that carbon dioxide, CO2, has a much greater density than air. In fact,
CO2 gas can displace air, which is why there are regulations in place that limit the
amount of dry ice allowed in elevators. In other words, do not get trapped in an
elevator, or any enclosed space, with someone who is transporting dry ice. Calculate
the pressure exerted by the CO2 gas, in atm, if the density was measured to be 1.983
g/L on a day where the temperature is 22.165 °C.
Answer: The pressure exerted by the [tex]CO_2[/tex] gas, in atm is 1.092
Explanation:
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = ?
V= Volume of the gas
T= Temperature of the gas = [tex]22.165^0C=(273+22.165)K=295.165K[/tex] (0°C = 273 K)
n= moles of gas = [tex]\frac{\text {given mass}}{\text {Molar mass}}[/tex]
R= Value of gas constant = 0.0821 Latm/K mol
[tex]P=\frac{mRT}{MV}[/tex] as [tex]Density=\frac{mass}{Volume}[/tex]
[tex]P=\frac{dRT}{M}[/tex] where d is density
[tex]P=\frac{1.983g/L\times 0.0821Latm/Kmol\times 295.165K}{44g/mol}=1.092atm[/tex]
Thus pressure exerted by the [tex]CO_2[/tex] gas, in atm is 1.092
What are anti-oxidants? Why are they added to fat and oil containing food?
Answer:
A substance that protects cells from the damage caused by free radicals (unstable molecules made by the process of oxidation during normal metabolism). Free radicals may play a part in cancer, heart disease, stroke, and other diseases of aging.
How many of the planets have an orbital period of less than one Earth year?
Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.
These planets are Mercury and Venus
Air movement and weather conditions are influenced by
A The Moon
B. Altitude
C. Thermal Energy
Answer:
The moon
Explanation:
The answer is A; Sorry i put the wrong answer by accident. I was trying to get to another question.
Give the answer below brainliest!!!
A. The moon influences the air movement
Identify the specific element that corresponds to each of the following electron configuration 1s2, 2s2, 2p6,3s2, 3p6
Explanation:
It's argon
As argon has atomic number 18 so it's configuration is
1s2, 2s2, 2p6,3s2, 3p6
Hope it will help :)❤
The given electron configuration 1s2, 2s2, 2p6, 3s2, 3p6 corresponds to the element argon (Ar).
Each number and letter in the electron configuration corresponds to a specific electron orbital and the number of electrons present in that orbital.
In this case, the electron configuration indicates that:
The first shell (n=1) has 2 electrons in the 1s orbital.
The second shell (n=2) has 2 electrons in the 2s orbital and 6 electrons in the 2p orbital.
The third shell (n=3) has 2 electrons in the 3s orbital and 6 electrons in the 3p orbital.
By adding up the total number of electrons (2 + 2 + 6 + 2 + 6), we find that the configuration represents 18 electrons. The element that has 18 electrons is argon, which has an atomic number of 18.
Therefore, the electron configuration 1s2, 2s2, 2p6, 3s2, 3p6 corresponds to the element argon (Ar).
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orbital hauny dump bell shape is...
a Sorbital il p-orbital 1 ild-orbital pulf orbital
Answer: the orbital with a hauny dump bell shape is the p- orbital.
Explanation:
An orbital can be defined as part of the nucleus of an atom that consists of an electron of maximum given energy. The shape of this part gives the shape of the orbitals. There are different types of orbitals which include:
--> s-orbitals: This is the type of orbital in which the probability of finding electron is the same in all directions usually at a given distance form the nucleus. Therefore the s-orbitals are SPHERICAL in shape about the nucleus and are non- directional.
--> d-orbitals: This is said to have a clover shape. This is because the electron is pushed out four times during the rotation when an opposite spin proton aligns gluons with three spin-aligned protons.
--> p-orbitals: This is said to have a DUMP-BELL shape. This is because there are three similar p-orbitals which are arranged mutually at right angles to each other along x,y, and z axes. All the three p-orbitals are of the same energy level.
A gas system has volume, moles and temperature of 1.52 L, 0.52 moles and 203K, respectively. What is the pressure in atm?
Answer:
P = 5.70 atm
Explanation:
Use PV = nRT
In this case, R = 0.0821 L·atm/mol·K (because we are finding the pressure in atm)P(1.52 L) = (0.52 moles)(0.0821)(203 K)
P = 5.701628947 atm
What is the coldest temperature ever achieved?
Answer:
The lowest natural temperature ever directly recorded at ground level on Earth is −89.2 °C
-273.144 degrees celsius. This is close to 'absolute zero', which is equal to -273.15 degrees celsius
Explanation: