________________________________
Using ideal gas law–
We'll use formula,
PV = nRT
Where,
P, Pressure = 1.40 atm
V, Volume = 42.4 L
n, Number of moles = ( To find ) ?
T, temperature = 25°C
= 25 + 273= 298 KelvinR, Ideal gas constant = 0.08206 L . atm/k. Mol
________________________________
Here, PV = nRT
Or, PV / RT = n
Or, [tex] \frac{1.40 \: atm \times42.4 \: litres }{298 \: k \: \times 0.08206 \: l.atm /k \: .mol} = n[/tex]
Or, n = 2.43 Moles.
Or, n = 2.43 Moles.
________________________________
I had to take help from my elder brother.
We tried our best, hope it helps you.
By– Debasis ( feat – Apollo)
________________________________
9.When two protons combine to form one nucleus, the result is one proton and one
neutron. How did the one proton become a neutron?
Answer:
josephmelichar
Explanation:
In the first step, 2 protons combine. In doing so one of the protons will convert into a neutron by losing a positron (positive electron) and a neutrino; this is also known as beta decay. This changes one of the protons (positive charge) into a neutron (neutral charge) and allows the two particles to combine.
Someone please help anyone who steals points will be reported
Answer:
ecosystem i think
Question 6
4 pts
6) Chromium crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell
is 289 pm, what is the density (in g/cm3) of chromium?
Show your work on a paper and upload it here.
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Answer:
7.15g/cm³
Explanation:
To solve this question we must know that a body-centered cubic unit cell contains 2 atoms.
The volume of the unit cell is:
289pm = (289x10⁻¹²m)³ =
2.414x10⁻²⁹m³ * (1cm³ / 1x10⁻⁶m³) = 2.414x10⁻²³cm³
And the mass is -Molar mass Chromium = 51.9961g/mol:
2atoms * (1mol / 6.022x10²³atoms) * (51.9961g / mol) =
1.727x10⁻²²g
The density is:
1.727x10⁻²²g / 2.414x10⁻²³cm³ =
7.15g/cm³What is the molarity of a solution that contains 152 g NaCl in 4.00 L solution?
Answer:
200.0lg
Explanation:
please give a brainliest
A pure substance is:___.
a. composed of two or more different types of atoms or molecules combined in variable proportions.
b. composed of two or more regions with different compositions.
c. composed of only one type of atom or molecule.
d. composed of two or more different types of atoms or molecules that has constant composition.
e. none of the above.
Answer:
c. composed of only one type of atom or molecule.
Explanation:
Pure substances are the substances that could be made up of only one type of particles and also it contains fixed structure. They would be further categorized as the elements & compounds. It is an elements that only consist of only one type of atom or molecule
Therefore as per the given situation, the option c is the answer
Answer: correct answer is C
Explanation:
How many sigma bond and pi bond are present
In volcanic areas, groundwater heated by magma is a source of
a.
lava flows.
b.
silica.
c.
geothermal energy.
d.
pyroclastic flows.
Answer:I believe it is c.
Explanation:
Answer:
C., or "geothermal energy."
What is the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3? (5 points) Group of answer choices 203 grams 161 grams 214 grams 187 grams
Answer:
203 grams
Explanation:
The no. of moles of (6.3 x 10²⁴ molecules--Avagadros number) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.
The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.
The answer is the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3 is 203 grams.
What is a molecule ?A molecule is a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance.
6.02214076 × 10²³ molecules constitutes 1 mole of NH₃
7.20 x 10²⁴ molecules constitutes of
[tex]\rm\dfrac{7.20 \times 10^{24}}{6.02\times 10^{23}}\\[/tex]
=11.19 moles
1 mole of NH₃ = 17 gm
11.19 mole will be = 17 * 11.19
= 203 grams
Therefore the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3 is 203 grams.
To know more about molecule
https://brainly.com/question/19922822
#SPJ2
In each pair of compounds, pick the one with the higher vapor pressure at a given temperature. Explain your reasoning.
a. CH4 or CH3CI
b. CH3CH2CH2OH or CH3OH
c. CH3OH or H2CO
Answer:
The answer is:
[tex]PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO[/tex]
Explanation:
Parts A:
The vapor pressure is higher in [tex]CH_4[/tex] because it is non-polar, while [tex]CH_3Cl[/tex] is polar. [tex]CH_4[/tex] has a lower molar weight as well.
Part B:
Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of [tex]CH_30H[/tex].
Part C:
[tex]H_2CO[/tex] does not show hydrogen due to the increased vapor pressure
[tex]CH_3OH[/tex] bonding.
When liquid water freezes into solid ice in the freezer,
Answer:
What about it? I don't get the question
What is the usable form of nitrogen that is found in the ground?
The exhaust from car engines spreads throughout our atmosphere and adds to pollution. What is this an example of?
A.
An increase in enthalpy
B.
An increase in thrust
C.
An increase in thermodynamics
D.
An increase in entropy
(D is the correct answer!)
Answer: When exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
Explanation:
A degree of randomness in the molecules of a substance is called entropy.
When exhaust from car engines spreads throughout our atmosphere then it means the molecules are moving at a faster speed due to which they spread into the atmosphere rapidly.
This means that the entropy of exhaust is increasing.
Thus, we can conclude that when exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
An increase in entropy
What element in group 1 is not considered an alkali metal?
A. Hydrogen (H)
B. Lithium (Li)
C. Potassium (K)
D. Cesium (Cs)
Answer:
A
Explanation:
Hydrogen isn't an alkali metal
Answer:
hydrogen
Explanation:
hydrogen is a gas not a metal
How much energy is contained in the six-cookie serving size recommended on the label?
Monounsaturated Fat 1g
The following data were obtained by gas-liquid chromatography on a 40-cm packed column:
Compound tR, min W, min
Air 1.9 —
Methylcyclohexane 10 0.76
Methylcyclohexene 10.9 0.82
Toluene 13.4 1.06
Calculate:
a. an average number of plates from the data.
b. the standard deviation for the average in (a).
c. an average plate height for the column.
Answer:
a)
For Methylcyclohexane N₁ = 2770
For Methylcyclohexene N₂ = 2827
For Toluene N₃ = 2557
b) the standard deviation for the average in (a) is 143.31
c)
For Methylcyclohexane; H₁ = 0.01444 cm
For Methylcyclohexene; H₂ = 0.01415 cm
For Toluene; H₃ = 0.01564 cm
Explanation:
Given the data in the question;
Gas-liquid chromatography on a 40-cm packed column:
Compound tR, min W, min
Air 1.9 —
Methylcyclohexane 10.0 0.76
Methylcyclohexene 10.9 0.82
Toluene 13.4 1.06
a)
an average number of plates from the data;
To get the Number of plates N, we use the following expression;
N = 16( tR / W )², we use it for Methylcyclohexane, Methylcyclohexene and Toluene
-
For Methylcyclohexane N₁ = 16( 10 / 0.76 )² = 16( 173.13 ) = 2770.08 ≈ 2770
-
For Methylcyclohexene N₂ = 16( 10.9 / 0.82 )² = 16( 176.7 ) = 2827.2 ≈ 2827
-
For Toluene N₃ = 16( 13.4 / 1.06 )² = 16( 159.8078 ) = 2556.9 ≈ 2557
b) standard deviation for the average in (a).
First we get the mean;
Mean N" = ( N₁ + N₂ + N₃ ) / 3 = ( 2770 + 2827 + 2557 ) / 3 = 8154 / 3 = 2718
Next we determine the deviation
d₁² = (N₁ - N")² = (2770 - 2718)² = (52)² = 2704
d₂² = (N₂ - N")² = (2827 - 2718)² = (109)² = 11881
d₃² = (N₃ - N")² = (2557 - 2718)² = (-161)² = 25,921
∴ ∑d²[tex]_i[/tex] = 40506
Standard Deviation S = √( ∑d²[tex]_i[/tex] / ( n-1 ) )
Standard Deviation S = √( 40506 / ( 3-1 )
Standard Deviation S = √( 40506 / 2 )
Standard Deviation S = √( 20253 )
Standard Deviation S = 143.31
Therefore, the standard deviation for the average in (a) is 143.31
c)
an average plate height for the column
Given that; Gas-liquid chromatography on a 40-cm packed column, L = 40 cm
-
For Methylcyclohexane; H₁ = L/N₁ = 40 / 2770 = 0.01444 cm
-
For Methylcyclohexene; H₂ = L/N₂ = 40 / 2827 = 0.01415 cm
-
For Toluene; H₃ = L/N₃ = 40 / 2557 = 0.01564 cm
A gas mixture contains each of the following gases at the indicated partial pressures: N2, 231 torr ; O2, 101 torr; and He, 155 torr .
What is the total pressure of the mixture?
What mass of each gas is present in a 1.00 -L sample of this mixture at 25.0 degrees Celsius?
Enter your answers numerically separated by commas.
Answer:
P = 487 Torr
mN₂ = 0.347 g
mO₂ = 0.174 g
mHe = 0.0333 g
Explanation:
Step 1: Calculate the total pressure of the mixture
The total pressure of the mixture (P) is equal to the sum of the partial pressures of the gases.
P = pN₂ + pO₂ + pHe
P = 231 Torr + 101 Torr + 155 Torr = 487 Torr
Step 2: Calculate the moles of each gas
We have 3 gases in a 1.00 L container at 25.0 °C (298.2 K). We can calculate the number of moles using the ideal gas equation: P × V = n × R × T.
nN₂ = pN₂ × V / R × T
nN₂ = 231 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.0124 mol
nO₂ = pO₂ × V / R × T
nO₂ = 101 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00543 mol
nHe = pHe × V / R × T
nHe = 155 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00833 mol
Step 3: Calculate the mass of each gas
The molar mass of N₂ is 28.01 g/mol.
mN₂ = 0.0124 mol × 28.01 g/mol = 0.347 g
The molar mass of O₂ is 32.00 g/mol.
mO₂ = 0.00543 mol × 32.00 g/mol = 0.174 g
The molar mass of He is 4.00 g/mol.
mHe = 0.00833 mol × 4.00 g/mol = 0.0333 g
Sample A: 300 mL of 1M sodium chloride
Sample B: 145 mL of 1.5 M sodium chloride
Which sample contains the larger concentration of sodium chloride?
Answer:
Sample A
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles (mol) ÷ volume (vol)
For Sample A:
V = 300ml = 300/1000 = 0.3 L
Molarity = 1M
n = number of moles (mol)
1 = n/0.3
n = 0.3moles
For Sample B:
V = 145 mL = 145/1000 = 0.145L
Molarity = 1.5 M
n = number of moles
1.5 = n/0.145
n = 1.5 × 0.145
n = 0.22 moles
Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.
What is matter?
O A. Anything that has energy and motion
B. Anything that takes up space and has mass
C. Anything that can be seen
D. Anything that can be measured
Answer:
B.Anything that takes up space and has mass
. Apply a coat of ____ to keep a wooden ladder structurally sound and functioning well. A. metallic paint B. oil-based paint C. linseed oil D.
Answer:
oil based paint
Explanation:
Give an example to illustrate the difference between "paying more in taxes ” and paying a higher tax rate .
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
Answer:
50 g Sucrose
Explanation:
Step 1: Given data
Mass of solution: 2000 gConcentration of the solution: 2.5%Step 2: Calculate the mass of sucrose needed to prepare the solution
The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:
2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose
Using stoichiometry, you predict that you should be able to use 314.0 g of Al to produce 1551 g of AlCi3. In your lab
exercise you actually produced 1400.0 g of aluminum chloride. What is your % yield for this reaction?
CORRECT ANSWER IS: 90.26% but what are the steps on how to get this answer ?
Answer:
90.26%
Explanation:
From the question given above, the following data were obtained:
Theoretical yield of AlCl₃ = 1551 g
Actual yield of AlCl₃ = 1400 g
Percentage yield =?
The percentage yield of the reaction can be obtained as follow:
Percentage yield = Actual yield / Theoretical yield × 100
Percentage yield = 1400 / 1551 × 100
Percentage yield = 140000 / 1551
Percentage yield = 90.26%
Thus, the percentage yield of the reaction is 90.26%
Civic participation is best described as the act of
avoiding harming other community members.
meeting one's legal responsibilities.
voting in every type of election.
going above and beyond one's legal duties.
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
meeting ones legal responsibility
A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.
Answer:
Explanation:
The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT
ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution
Kb for water = .51°C /m
moles of sugar = 16.90 / 342.3
= .04937 moles
m = moles of sugar / kg of water
= .04937 / .04090
= 1.207
ΔT = Kb x m
= .51 x 1.207
= .62°C .
So , boiling point of water = 100.62°C .
What does a velocity measurement include that a speed measurement does not?
time
direction
distance
acceleration
If a medicine ball in a gym has a mass of 4.0 Kg, what is its weight?
Answer:
Weight = 0.4Explanation:
Given Information :
Mass = 4.0kg
Acceleration due to gravity = 10 m/s
Weight= ?
[tex]Weight = \frac{mass}{acceleration\: due \:to \:gravity} \\\\W= \frac{4.0}{10} \\\\W= \frac{2}{5} \\\\W=0.4[/tex]
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the same empirical formula as propene, C3H6
Sample Mass of carbon Mass of hydrogen
A 60g 12g
B 72g 12g
C 84g 10g
D 90g 10g
a. Sample A
b. Sample B
c. Sample C
d. Sample D
Answer:
Sample B
Explanation:
In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.
Let's do Sample A:
C: 60 g; H: 12 g
1. Calculate moles:
We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):
C: 60 / 12 = 5
H: 12 / 1 = 12
2. Determine number of atoms in the formula
In this case, we just divide the lowest moles obtained in the previous part, by all the moles:
C: 5 / 5 = 1
H: 12 / 5 = 2.4 or rounded to two
3. Write the empirical formula:
Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:
C₁H₂ = CH₂
Therefore, sample A is not the same as propene.
Sample B:
C: 72 g H: 12 g
Following the same steps, let's determine the empirical formula for this sample
C: 72 / 12 = 6 ---> 6 / 6 = 1
H: 12 / 1 = 12 ----> 12 / 6 = 2
EF: CH₂
Sample C:
C: 84 g H: 10 g
C: 84 / 12 = 7 ----> 7 / 7 = 1
H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1
EF: CH
Sample D
C: 90 g H: 10 g
C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1
H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1
EF: CH
Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:
C3H6 -----> CH₂
Therefore, sample B is the one that match completely. Sample B would be the one.
Hope this helps
a sample of mg(hco3)2 contains 1.8 moles of oxygen atom find the number of carbon atoms in the given sample
please solve fast very much urgent
Answer:
0.60 moles of atoms of carbon
Explanation:
Step 1: Given data
Chemical formula of the compound: Mg(HCO₃)₂Moles of oxygen atoms: 1.8 molesStep 2: Calculate the number of carbon atoms in the given sample
According to the chemical formula of the compound, the molar ratio of C to O is 2:6, that is, there are 2 moles of atoms of C every 6 moles of atoms of O. The number of moles of atoms of C is:
1.8 mol O × 2 mol C / 6 mol O = 0.60 mol C
Which of the following is the BEST description of a budget? *
10
Answer:
A budget is an estimation of revenue and expenses over a specified future period of time and is usually compiled and re-evaluated on a periodic basis. Budgets can be made for a person, a group of people, a business, a government, or just about anything else that makes and spends money.
The elements in the periodic table are not always represented by
the first one or two letters in their names. What are some examples of
this (list three)?
Unnillium (101)
Unnilbium (102)
Unniltrium (103)
Unnilquadium (104)
Unnilpentium (105)
Unnilhexium (106)
Unnilseptium (107)
Unniloctium (108)
Unnilennium (109)
Ununnillium (110)