Answer:
The 3rd answer down.
Na²O (sodium oxide) will be a base when exposed to water H²O
Explanation:
Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).
The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.
Calculate the molarity of a 17.5% (by mass) aqueous solution of nitric acid. Select one: a. 2.74 m b. 4.33 m c. 0.274 m d. 3.04 m e. The density of the solution is needed to solve the problem.
Answer:
Option e.
Explanation:
Molarity is the concentration that indicates moles of solute in 1 L of solution.
We have another concentration, percent by mass.
Percent by mass indicates mass of solute in 100 g of solution.
Our solute is HNO₃, our solvent is water.
17.5 g of nitric acid is the mass of solute. We can convert them to moles:
17.5 g . 1mol / 63g = 0.278 moles
We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.
Density = Mass / Volume
Mass / Density = Volume
Once we have the volume, we need to be sure the units is in L, to determine molarity
M = mol /L
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
Convert 3.00 x 10^21 atoms of copper to moles.
Convert 2.25 x 10^18 molecules of carbon dioxide to moles.
Answer:
1) 0.00498 mol Cu.
2) 0.00000374 mol CO₂
Explanation:
Question 1)
We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.
Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:
[tex]\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}[/tex]
We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:
[tex]\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}[/tex]
Cancel like terms:
[tex]=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }[/tex]
Simplify:
[tex]\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}[/tex]
Use a calculator:
[tex]= 0.004981... \text{ mol Cu}[/tex]
Since the resulting answer must have three significant figures:
[tex]= 0.00498\text{ mol Cu}[/tex]
So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.
Question 2)
We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.
By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}[/tex]
To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:
[tex]\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}[/tex]
Cancel like terms:
[tex]\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}[/tex]
Simplify:
[tex]\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}[/tex]
Use a calculator:
[tex]=0.000003736...\text{ mol CO$_2$}[/tex]
Since the resulting answer must have three significant figures:
[tex]= 0.00000374\text{ mol CO$_2$}[/tex]
So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.
Answer:
Explanation:
by definition, 1 mole contains 6.02 x 10^23 of atoms (for elements) or molecules (for compounds)
3.00 x 10^21 atoms of copper / 6.02 x 10^23 of atoms
= 0.004983 moles of copper
= 4.98 x 10^(-3) moles of copper
2.25 x 10^18 molecules of carbon dioxide / 6.02 x 10^23 of molecules
= 0.000003737 moles of carbon dioxide
= 3.74 x 10^(-6) moles of carbon dioxide
Gaseous BF3 and BCl3 are mixed in equal molar amounts. All B-F bonds have about the same bond enthalpy, as do all B-Cl bonds. Compare the numbers of microstates to explain why the mixture tends to react to form BF2Cl(g) and BCl2F(g
Solution :
[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]
Explanation 1 :
Spontaneity of the reaction is based on two factors :
-- the tendency to acquire a state of minimum energy
-- the energy of a system to acquire a maximum randomness.
Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.
Explanation 2 :
A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].
It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.
How many atoms are present in 0.45 moles of P4010
Answer:
80g
Explanation:
mass oxygen present in 1 mole of p4010
16×10=160gm
similarly
for 0.5 moles of p4010 160/2= 80gm
The number of atoms present in 0.45 moles of P₄O₁₀ is 1.08 x 10²³ atoms.
To determine the number of atoms, we use Avogadro's number, which states that there are approximately 6.022 x 10²³ particles (atoms, molecules, or formula units) in one mole of a substance.
In this case, we are given 0.45 moles of P₄O₁₀. To calculate the number of atoms, we multiply the number of moles by Avogadro's number:
Number of atoms = 0.45 moles P₄O₁₀ x (6.022 x 10²³ atoms / 1 mole)
Number of atoms = 2.7139 x 10²³ atoms
Rounding to three significant figures, the number of atoms present in 0.45 moles of P₄O₁₀ is approximately 1.08 x 10²³ atoms.
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The density of mercury is 13.6 g/cm3, What is its density in mg/mm3?
Answer:
Density of mercury is 13600 kg
A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase
Answer:
the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Explanation:
Given the data in the question;
Co = 53 or [ 53 wt% B-47 wt% A ]
W∝ = 0.5 = Wβ
Cβ = 92 or [ 92 wt% B-8 wt% A ]
Now, lets set up the Lever rule for W∝ as follows;
W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]
so we substitute our given values into the expression;
0.5 = [ 92 - 53 ] / [ 92 - C∝ ]
0.5 = 39 / [ 92 - C∝ ]
0.5[ 92 - C∝ ] = 39
46 - 0.5C∝ = 39
0.5C∝ = 46 - 39
0.5C∝ = 7
C∝ = 7 / 0.5
C∝ = 14 or [ 14 wt% B-86 wt% A ]
Therefore, the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Classify each molecule as an alcohol, ketone, or aldehyde based on its name. Propanone (acetone) Choose... Ethanal Choose... 3-phenyl-2-propenal Choose... Butanone Choose... Ethanol Choose... 2-propanol Choose...
Answer:
1.) Propanone (ketone)
2.) Ethanal( aldehyde)
3.) 3-phenyl-2-propenal (aldehyde)
4.) Butanone (ketone)
5.) Ethanol ( alcohol)
6.) 2-propanol (alcohol)
Explanation:
In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.
Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.
Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal
When 250. mg of eugenol, the molecular compound responsible for the odor of oil of cloves, was added to 100. g of camphor, it lowered the freezing point of camphor by 0.62 8C. Calculate the molar mass of eugenol.
Answer:
Molar mass for eugenol is 161.3 g/mol
Explanation:
This question talks about freezing point depression:
Our solute is eugenol.
Our solvent is camphor.
Formula to state the freezing point depression difference is:
ΔT = Kf . m . i where
ΔT = Freezing T° of pure solvent - Freezing T° of solution
In this case ΔT = 0.62°C
Kf for camphor is: 37°C /m
As eugenol is an organic compund, i = 1. No ions are formed.
To state the molar mass, we need m (molal)
Molal are the moles of solute in 1kg of solvent. Let's replace data:
0.62°C = 40 °C/m . m . 1
0.62°C / 40 m/°C = 0.0155 m
We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg
0.0155 molal = moles of solute / 0.1 kg
0.0155 m/kg . 0.1 kg = 0.00155 moles
We know that these moles are contained in 250 mg, so the molar mass will be:
0.25 g / 0.00155 mol = 161.3 g/mol
Notice, we convert mg to g, for the units!
What force is behind us when we ride a bike?
Answer:
gravity, ground, friction, rolling resistance, and air resistance.
Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.
In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?
For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)
Answer:
Mass of C = 47.37g
Mass of H = 10.59g
Mass of O = 42.04g
The total mass of these elements is 100g, taking a proportion of their molar masses.
C = 47.37/12= 3.95
H = 10.59/1 = 10.59
O = 42.04/16= 2.63.
Dividing through with the smallest proportion which is 2.63
C=3.95/2.63 = 1.5
H =10.59/2.63 =4
O = 2.63/2.63= 1
Multiplying through by 2 to get a whole number.
C = 1.5x2 = 3
H= 4x2 = 8
O = 1x2= 2
The empirical formula is C3H6O2
(Empirical formula)n= molecular mass
(C3H8O2)n =228.276
(12x3 +8+16x2)n= 228.276
76n = 228.276
n = 228.276/76
n = 3
Molecular formula = Empirical formula
=(C3H8O2)3 = C9H24O6
The molecular formula is C9H24O6
once formed, how are coordinate covalent bonds different from other covalent bonds?
Answer:
[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]
Explanation:
A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).
Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.
Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.
Which intermolecular force plays a pivotal role in biological molecules such as proteins and DNA ?
•hydrogen bonding
•dispersion force
•dipole-dipole force
•Ion-dipole force
Answer:
hydrogen bonding
Explanation:
just took the test :D
In the given range,at what temperature does oxy gen have the highest solubility?
A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]
By rearrangement:
[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]
Multiplying through by (-) and solving the quadratic equation:
[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]
[tex](-0.00015 + x) (0.000168 + x) =0[/tex]
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]
= 10.71%
Which particle has a mass of 9.11 x 10^-28g and charge of -1?
A. electron
B. proton
C. neutron
QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?
OPTIONS:-
A. electron
B. proton
C. neutron
ANSWER:-
CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER
THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER
MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]
CHARGE ON ELECTRON:- [tex] -1[/tex]
SO URE ANSWER IS ELECTRON
What is the specific rotation of 13g of a molecule dissolved in 10 mL of solvent that gives an observed rotation of 23 degrees in a sample tube of 10 cm.
Answer:
[tex]\alpha=17.7[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=13g[/tex]
Volume [tex]V=10mL[/tex]
Angle [tex]\theta=23[/tex]
Sample Tube=10cm
Generally the equation for concentration is mathematically given by
[tex]C=m/v[/tex]
[tex]C=\frac{13}{10}\\C=1.3g/mL[/tex]
Therefore the Specific Rotation
[tex]\alpha=frac{\theta }{m*l}[/tex]
[tex]\alpha=frac{23 }{1.3*1.0}[/tex]
[tex]\alpha=17.7[/tex]
What is the energy change when 78.0 g of Hg melt at −38.8°C
Answer:
The correct answer is - 2.557 KJ
Explanation:
In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.
we can calculate this energy by the following formula:
Q = met
where, m = mass,
e = specific heat
t = temperature
then,
Q = 78*0.14* (273-38.8)
here 0.14 = C(Hg)
= 2.557 Kj
How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma
Answer:
C. By super-cooling certain types of plasma.
Explanation:
Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.
Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.
When certain types of plasma are super cooled, Bose-Einstein condensate are formed.
Identify the phase of the copper product after each reaction in the copper cycle.
The addition of HNO3 HNOX3 to Cu ______________
The addition of H2SO4 HX2SOX4 to CuO ____________ The addition of Z n Zn to C u S O 4 CuSOX4 Choose... The addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 Choose... The heating of C u ( O H )
Answer:
addition of HNO3 HNOX3 to Cu - Aueous
addition of H2SO4 HX2SOX4 to CuO - Aqueous
addition of Z n Zn to C u S O 4 CuSOX4 - Solid
addition of N a O H NaOH to C u ( N O 3 ) 2 Cu(NOX3)X2 - Solid
heating of C u ( O H ) - Solid
Explanation:
Copper when introduced with acids form an aqueous solution and fumes are released in air during the chemical reaction. When NaOH is added to copper then solid copper product is released. Copper dissolves on HNO but does not dissolves in HCL.
Consider the reaction “2 SO2 (g) + O2 (g) = 2 SO3 was 0.175 M. After 50 s the concentration of SO2 Date: (g)”. Initial concentration of SO2 (g) (g) became 0.0500 M. Calculate rate of the reaction
Answer:
The answer is "[tex]1.25 \times 10^{-3} \ \frac{m}{s}[/tex]"
Explanation:
Calculating the rate of the equation:
[tex]=-\frac{1}{2} \frac{\Delta [SO_2]}{\Delta t} =-\frac{\Delta [O_2]}{\Delta t}= +\frac{1}{2} \frac{\Delta [SO_3]}{\Delta t}\\\\=\frac{\Delta [SO_2]}{\Delta t}=\frac{0.0500-0.175\ M}{505}= -2.5 \times 10^{-3} \ \frac{m}{s}\\\\[/tex]
Rate:
[tex]=\frac{-2.5 \times 10^{-3}}{2}=1.25 \times 10^{-3} \ \frac{m}{s}[/tex]
Consider the following reaction:
CO(g)+2H2(g)⇌CH3OH(g)
A reaction mixture in a 5.15-L flask at a certain temperature initially contains 26.6 g CO and 2.36 g H2. At equilibrium, the flask contains 8.63 g CH3OH.
Part A
Calculate the equilibrium constant (Kc) for the reaction at this temperature.
Answer:
26.6
Explanation:
Step 1: Calculate the molar concentrations
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M
[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M
[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M
Step 2: Make an ICE chart
CO(g) + 2 H₂(g) ⇄ CH₃OH(g)
I 0.184 0.227 0
C -x -2x +x
E 0.184-x 0.227-2x x
Since [CH₃OH]e = x, x = 0.0523
Step 3: Calculate all the concentrations at equilibrium
[CO]e = 0.184-x = 0.132 M
[H₂]e = 0.227-2x = 0.122 M
[CH₃OH]e = 0.0523 M
Step 4: Calculate the equilibrium constant (Kc)
Kc = [CH₃OH] / [CO] [H₂]²
Kc = 0.0523 / 0.132 × 0.122² = 26.6
How can this product be achieved using the starting material shown?
Answer:
this product can be achieved using the starting material shown is by use of NaOH as catalyst.
Answer:
By using NaOH as catalyst.
Explanation:
This product can be achieved using the starting material shown is by the use of the NaOH as catalyst.
What is an emission spectrum?
A. The total amount of energy emitted by an element
B. The products created when an element is burned
C. The energy absorbed when an electron gains energy
D. The colors of light given off when an element loses energy
Answer:
D
Explanation:
The electromagnetic radiation is emitted due to a particle moves from a higher to a lower energy state
An emission spectrum is the colors of light given off when an element loses energy. Therefore, option D is correct.
What is emission spectrum ?The electromagnetic radiation spectrum produced when an electron changes from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound.
An emission spectrum is the range of radiations that are released in different places when electrons jump back and forth between higher and lower energy levels to achieve stability.
Since what you are seeing is the direct radiation produced by the source, this form of spectrum is also known as an emission spectrum. You can see all the colors in the Sun's spectrum because light from the Sun is produced at practically all energies in the visible spectrum.
Thus, option D is correct.
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What direction would equilibrium moves towards based on the following if we increased the volume of the container.
[tex]2A_{(g)} + 5B_{(g)} + 12C_{(g)}[/tex] ↔ [tex]14AC_{(g)} + 5B_{(s)}[/tex]
Answer choices:
a) reactants
b) no change
c) products
d) decrease in volume
Please help!
To answer this question, we will first find out the number of gaseous moles on each side of the equilibrium
on the left:
we have 2 moles of A, 5 moles of B and 12 moles of C
which gives us a grand total of 19 gaseous moles
on the right:
here, we have 14 moles of AC gas, we will not count the number of moles of B because it's a solid
giving us 14 gaseous moles on the right
Where does the reaction shift?
more gaseous moles means more space taken, because gas likes to fill all the space it can
if we have more volume, more gas can move around without colliding (reacting) with each other
Hence more volume favors the side with more gaseous moles
here, the left has more gaseous moles. So we can say that the reaction will shift towards the left, or the reactants side
Answer:
Explanation:
given reversible chemical reaction:
2A(g) + 5B(g) + 12C(g) ↔ 14AC(g) + 5B(s)
chemicals in solid form do not take up a lot of volume so change in container volume has no effect
look at chemicals in gas form only:
the total no. of moles of reactants in gas form = 2 + 5 + 12 = 19
the total no. of moles of products in gas form = 14
so an increase in volume of the container will favor the reaction direction with higher volume n high volume means higher no. of moles
the ans is the equilibrium will move towards a) reactants
Question In nickel-cadmium batteries: Select the correct answer below: the anodes are nickel-plated and the cathodes are cadmium-plated the anodes are cadmium-plated and the cathodes are nickel-plated both the anodes and cathodes are plated with a nickel-cadmium alloy none of the above
Answer:
the anodes are cadmium-plated and the cathodes are nickel-plated
Explanation:
Nickel cadmium battery works on the principle as by the other cell. There is anode and a cathode which is separated by a separator (spiral shaped inside the case). The anode is negative and is cadmium plated while the cathode is positive and is nickel plated. An electrolyte is also used.
So the correct answer is : "The anodes are cadmium-plated and the cathodes are nickel-plated."
Generally the vapor pressure of a liquid is related to: I. the amount of liquid II. atmospheric pressure III. temperature IV. intermolecular forces
Answer:
Atmospheric pressure, because a liquid is said to be boiling when the vapour pressure equals the atmospheric pressure.
What is true of all matter?
A. It pushes or pulls on objects.
B. You can see it.
C. It gives off heat energy.
D. It has mass.
Read the chemical equation.
N2 + 3H2 - 2NH3
Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2
if all measurements are taken at the same temperature and pressure?
A 2.8 liters
B 3.2 liters
C 5.4 liters
D 6.3 liters
Answer:
A 2.8 liters
Explanation:
Step 1: Write the balanced equation
N₂ + 3 H₂ ⇄ 2 NH₃
Step 2: Establish the appropriate volume ratio
At the same temperature and pressure, the volume ratio of H₂ to NH₃ is 3:2.
Step 3: Calculate the volume of ammonia produced from 4.2 L of hydrogen
4.2 L H₂ × 2 L NH₃/3 L H₂ = 2.8 L
Determine the equilibrium constant, Keq, at 25°C for the reaction
2Br- (aq) + I2(s) <--> Br2(l) + 2I- (aq)
Eocell = (0.0257/n) lnKeq, Calculate Eocell from Use this equation to calculate K value.
Eo (I2/I-) = +0.53, Eo (Br2/Br-) = +1.07,
Explanation:
The given chemical reaction is:
[tex]2Br^- (aq) + I_2(s) <-> Br_2(l) + 2I^- (aq)[/tex]
[tex]E^ocell=oxidation potential of anode + reduction potential of cathode\\[/tex]
The relation between Eo cell and Keq is shown below:
[tex]deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell[/tex]
The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,
[tex]E^ocell= (-1.07+0.53)V\\=-0.54V[/tex]
F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:
[tex]ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3[/tex]
Answer:
Keq=6.13x10^33
A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.
Answer:
density of second liquid = 650 kg/m³
Explanation:
Given that:
The volume of the plastic block submerged inside the water = 0.5 V
The force on the plastic block = [tex]\rho_1V_1g[/tex]
[tex]= 0.5p_1 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
W [tex]= 0.5p_1 V_g[/tex]
[tex]\rho Vg = 0.5p_1 V_g[/tex]
[tex]\rho = 0.5 \rho _1[/tex]
where;
water density [tex]\rho _1[/tex] = 1000
[tex]\rho = 0.5 (1000)[/tex]
[tex]\rho = 500 kg/m^3[/tex]
In the second liquid, the volume of plastic block in the water = (100-23)%
= 77% = 0.7 V
The force on the plastic block is:
[tex]= 0.77p_2 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
[tex]W = 0.77p_2 V_g[/tex]
[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]
[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]
