How many unpaired electrons are in the following complex ions?
a. Ru(NH3)62+ (low-spin case)
b. Ni(H2O)62+
c. V(en)33+
Solution :
A complex ion may be defined as a metal ion that is located at the center with other molecules or ions surrounding the center ion.
In the context, the number of the unpaired electrons for the following complex ions are :
1. [tex]$Ru(NH_3)62+$[/tex] (low spin)
[tex]$Ru^{2+} = [Kr] \ 4d^6 \ 5s^0$[/tex]
Here all the d-electrons are paired. Therefore, there are no unpaired electrons.
2. [tex]$Ni(H_2O)62+$[/tex] (high spin)
[tex]$Ni^{2+} = [Ar] \ 4s^0 \ 3d^8$[/tex]
Therefore, it has two unpaired electrons.
3. [tex]$V(en)33+$[/tex] (high spin)
[tex]$V=[Ar] \ 3d^2 \ 4s^0$[/tex]
It has 2 unpaired electrons
n-hexane boils at 68.7 oC at 1.013 bar and the heat of Vaporization is 28850
J/mol. Calculate Its Entropy ?
What mass of sucrose is needed to make 300.0 mL of a 0.5 M solution? (molar mass=
342.34 g/mol)
A silver bar with a specific heat of 0.235J/gºC has a mass of 250.g and is heated from 22.0°C to 68.5°C. How much heat will the silver bar absorb?
Answer:
2731.875 J
Explanation:
Applying
Q = cm(t₂-t₁)................ Equation 1
Where Q = amount of heat, c = specific heat capacity of silver, m = mass of silver, t₂ = Final temperature of silver, t₁ = Initial temperature of silver
From the question,
Given: c = 0.235 J/g°C, m = 250 g, t₁ = 22.0°C, t₂ = 68.5°C
Substitute these values into equation 1
Q = 0.235×250(68.5-22.0)
Q = 58.75(46.5)
Q = 2731.875 J
How many grams of carbon are In 25 g of CO2?
Answer: almost 7g
Explanation: 25 grams of carbon dioxide contains 6.818 grams of carbon.
when 18.0 g H20 is mixed with 33.5 g Fe, which is the limiting reactant?
Answer:
si.mple fe
Explanation:
ghhsshzhzbbzhh
What is the pH of a 6.2 x 10-5 M NaOH solution?
Answer:
4.21Explanation:
The pH of a solution can be found by using the formula
[tex]pH = - log [ {H}^{+} ][/tex]
From the question we have
[tex]ph = - log(6.2 \times {10}^{ - 5} ) \\ = 4.207608[/tex]
We have the final answer as
4.21Hope this helps you
A 16.0 L sample contains 0.80 moles of a gas. If the gas is at the
same pressure and temperature, what is the number of moles when
the volume increases to 24.0 L? NOTE: You must show your
calculation on the attached scratch paper, including which of the
Gas Law formulas you used. *
A. 0.83 mol
B. 1.2 mol
C. 0.53 mol
D. 1.9 mol
(Please show your work)
[tex]\huge\bf\red{\underline{\underline{Answer:-}}}[/tex]
Here, 16 L = 0.80 Mole
Or, 1 L = ( 0.80 / 16 ) MoleOr, 1 L = 0.05 Mole_____________________________
Now, 24 L = 0.05 × 24 = [tex] \frac{5}{100} \times 24[/tex]= 120/100= 12/10= 1.2 Mole ( Ans )What causes the recognition of an element from its energy signature?
O
A. Each element emits radiations at random frequencies.
B. Each element emits energy that can be transformed into sound.
C. Each element emits electromagnetic waves that travel faster than light.
D. Each element emits electromagnetic waves at a unique set of wavelengths.
Answer:
I'm so sorry if this is wrong but I think its B
Atomic number: 19 is what Chemical symbol?
Answer:
oficialismo hemostasia Jalisco
Answer:
K
Explanation:
Potassium is atomic no 19 and it is k because is from kalium, the Meldiaeval Latin for potash, which may have derived from an arabic word quality, meaning alkali.
Atomic mass, 39.098amu
Oxidation status, +1
Year discovered, 1807
Melting point, 63.28 degrees Celsius(145.90degrees Fahrenheit )
Boiling point, 760 degrees Celsius (1400 degrees Fahrenheit )
At the start of a reaction, there are 0.0249 mol N2,
3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a
3.50 L reaction vessel at 375°C. If the equilibrium constant, K, for the reaction:
N2(g) + 3H2(g)= 2NH3(g)
is 1.2 at this temperature, decide whether the system is at equilibrium or not. If it is not, predict in which direction, the net reaction will proceed.
Answer:
Explanation:
The reaction is given as:
[tex]N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}[/tex]
The reaction quotient is:
[tex]Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
From the given information:
TO find each entity in the reaction quotient, we have:
[tex][NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}[/tex]
[tex][N_2] = \dfrac{0.024 }{3.5}[/tex]
[tex][N_2] = 0.006857[/tex]
[tex][H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}[/tex]
[tex][H_2] = 9.17 \times 10^{-3}[/tex]
∴
[tex]Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135[/tex]
However; given that:
[tex]K_c = 1.2[/tex]
By relating [tex]Q_c \ \ and \ \ K_c[/tex], we will realize that [tex]Q_c \ \ < \ \ K_c[/tex]
The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.
In 1773 Benjamin Franklin, investigating the calming effect of oil on the water of a pond near London, observed that a teaspoonful of oil would cover about half an acre of the surface. From this information, estimate the height or thickness of an oil molecule in the floating film of oil. Then estimate the cross sectional area of an oil molecule in the floating film You may assume that Franklin used a lamp oil, like camphene, C10H16.
Required:
Calculate the thickness (in nm) of the surface film and the surface area (in nm2) occupied by an oleic acid molecule on water.
Answer:
a. 2.44 nm
b. 15.33 nm²
Explanation:
a. Calculate the thickness (in nm) of the surface film
Since the volume of a teaspoon equals V = 4.93 cm³ and the area of the oil film is half an acre = 1/2 × 4046.86 m² = 2023.43 m²
The volume of the oil film, V' equals the volume of the oil in the teaspoon, V
V' = Ah where A = cross-sectional area of oil film = 2023.43 m² = 2023.43 × 10⁴ cm² and h = height of oil film
So, V' = V
Ah = V
h = V/A
= 4.93 cm³/2023.43 × 10⁴ cm²
= 0.00244 × 10⁻⁴ cm
= 2.44 × 10⁻⁷ cm
= 2.44 × 10⁻⁷ cm × 1m/100 cm
= 2.44 × 10⁻⁹ m
= 2.44 nm
b. The surface area (in nm2) occupied by an oleic acid molecule on water.
Since the height of the oil film equals the diameter of the oil molecule, and the molecule is assumed to be a sphere of radius, r. Its surface area is thus A = 4πr²
r = h/2 = 2.44 nm/2 = 1.22 nm
A = 4πr²
A = 4π(1.22 nm)²
A = 4.88π nm²
A = 15.33 nm²
chemical properties of citric acid
Answer:
Citric Acid is a weak acid with a chemical formula C6H8O7.
...
Properties of Citric Acid – C6H8O7.
C6H8O7 Citric Acid
Molecular Weight/ Molar Mass 192.124 g/mol
Density 1.66 g/cm³
Boiling Point 310 °C
Melting Point 153 °C
Hope it helps❤️
If a drop of red dye is placed in the middle of a shallow dish of water, the dye will diffuse slowly and a red cloud of dye will grow larger with time.
a. True
b. False
Answer:
True
Explanation:
Diffusion is the movement of the particles of a substances from a region of higher concentration to a region of lower concentration.
Diffusion occurs in response to a concentration gradient.
In this case, the dye is at a higher concentration at the middle of the shallow dish of water. Diffusion causes the dye to spread out gradually thereby causing the red cloud to grow larger with time.
what is an example of a change in genetic traits of an organism do to human affect
Answer:
A person's skin color, hair color, dimples, freckles, and blood type are all examples of genetic variations that can occur in a human population.
Explanation:
A 0.750 M sample of phosphorus pentachloride decomposes to give phosphorus trichloride and chlorine gases.If the equilibrium concentration of PCl5 is 0.650M,what is the equilibrium constant for the reaction?
PCl5(g) --------> PCl3(g)+ Cl₂(g)
A) Keq = 0.0133
B) Keq = 0.0154
C) Keq = 0.133
D) Keq = 0.154
E) Keq = 65.0
Answer:
Explanation:
Amount of phosphorus pentachloride reacted = 0.75 M - 0.65 M = 0.10 M
PCl₅(g) --------> PCl₃(g)+ Cl₂(g)
1 mole 1 mole 1 mole
Amount of phosphorus pentachloride reacted = 0.75 M - 0.65 M = 0.10 M
PCl₃(g) and Cl₂(g) formed are .10 M and .10 M
Keq = [ PCl₃] x [ Cl₂ ] / [ PCl₅ ]
= .10 M x .10 M / .65 M
= .0154
B ) is the correct choice .
0=4
Balance this equation
H₂sicl2+ H₂O → H8Si4O4 + HCl
Balanced Equation is
4H2SiCl2+4H2O → H8Si4O4 + 8HCl
how to solve x² in differential
Answer:
x² = mutiphy by them self
Explanation:
Enthalpy change is the
a. amount of energy absorbed or lost by a system as heat during a process at constant pressure.
b. entropy change of a system at constant pressure.
c. pressure change of a system at constant temperature.
d. temperature change of a system at constant pressure.
Answer:
I believe is A
Explanation:
Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure.
Please hurry!
Best answer gets Brainliest!!
Please answer the question ASAP
Scientists clarify an element based on what criteria
A)Atomic number
B)Mass number
C)Atomic mass
D)Atomic symbol
Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen,
and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50
Answer:
C₃H₈O
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.031 g
Mass of CO₂ = 2.265 g
Mass of H₂O = 1.236 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:
For carbon, C:
Mass of CO₂ = 2.265 g
Molar mass of CO₂ = 44.01 g/mol
Molar mass of C = 12.01 g/mol
Mass of C =?
Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂
Mass of C = 12.01/44.01 × 2.265
Mass of C = 0.618 g
For hydrogen, H:
Mass of H₂O = 1.236 g
Molar mass of H₂O = 18.02 g/mol
Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol
Mass of H =?
Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O
Mass of H = 2.02/18.02 × 1.236
Mass of H = 0.139 g
For oxygen, O:
Mass of compound = 1.031 g
Mass of C = 0.618 g
Mass of H = 0.139 g
Mass of O =?
Mass of O = Mass of compound – (mass of C + mass of H)
Mass of O = 1.031 – ( 0.618 + 0.139)
Mass of O = 1.031 – 0.757
Mass of O = 0.274 g
Finally, we shall determine the empirical formula. This can be obtained as follow:
C = 0.618 g
H = 0.139 g
O = 0.274 g
Divide by their molar mass
C = 0.618 / 12.01 = 0.051
H = 0.139 / 1.01 = 0.138
O = 0.274 / 16 = 0.017
Divide by the smallest
C = 0.051 / 0.017 = 3
H = 0.138 / 0.017 = 8
O = 0.017 / 0.017 = 1
Therefore, the empirical formula of the compound is C₃H₈O
Based on the data provided, the empirical formula of the compound is C₃H₈O
What is empirical formula?The empirical formula of compound is its simplest formula showing the mole ratio of the elements in the compound.
First, the mass of carbon, hydrogen and oxygen in the compound is determined first as follows:
For carbon, C:
Mass of CO₂ = 2.265 g
Molar mass of CO₂ = 44.01 g/mol
Molar mass of C = 12.01 g/mol
Mass of C = 12.01/44.01 × 2.265
Mass of C = 0.618 g
For hydrogen, H:
Mass of H₂O = 1.236 g
Molar mass of H₂O = 18.02 g/mol
Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol
Mass of H = 2.02/18.02 × 1.236
Mass of H = 0.139 g
For oxygen, O:
Mass of compound = 1.031 g
Mass of C = 0.618 g
Mass of H = 0.139 g
Mass of O = Mass of compound – (mass of C + mass of H)
Mass of O = 1.031 – ( 0.618 + 0.139)
Mass of O = 0.274 g
To determine the empirical formula of a compound, the mole ratio of the elements are determined as follows:
Mole ratio = reacting mass/molar massMole ratio of the elements:
Carbon Hydrogen Oxygen
0.618 g/12.01 0.139 / 1.01 0.274 / 16
0.051 0.138 0.017
Divide by the smallest ratio to convert to whole numbers
0.051 / 0.017 0.138 / 0.017 0.017 / 0.017
3 : 8 : 1
Therefore, the empirical formula of the compound is C₃H₈O
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3.05 moles of H2O is equivalent to (blank amount) molecules of water.
Show your work.
Answer:
[tex]1 \: mole \: = 6.02 \times {10}^{23} \: molecules \\ 3.05 \: moles \: = (3.05 \times 6.02 \times {10}^{23} ) \: molecules \\ = 1.8361 \times {10}^{24} \: molecules[/tex]
Chemical bond stretching frequencies depend on two major factors, namely atomic weight and bond stiffness. The frequency absorbed is directly proportional to the bond strength (stiffness) but inversely proportional to the atomic weight. Which of the following bonds will give the highest absorption frequency value?
A. C single bond C.
B. C double bond C.
C. All would be the same frequency.
D. C triple bond C.
Answer:
D. C triple bond C.
Explanation:
The absorption frequency value is highest for carbon-carbon triple bond. The stretching frequencies are much higher than the bending frequency. It mainly depends upon the strength of the bonds as well as masses of the bonded atoms.
The triple bonds have higher stretching strength than the double bonds and double bonds have higher strength than single carbon to carbon bonds.
The triple bonds has a higher bond order, i.e. 3. So it has maximum stretching bonds.
Which of the following statement is true?
a. The resonance effect of the hydroxyl group stabiizes the anionic intermediate.
b. The resonance effect of the hydroxyl group stabilizes the cationic
c. The inductive effect of the hydroxyl group stabiizes the cationic intermediate
d. The inductive effect of the hydroxyl group stabiizes the anionic intermediate
Answer:
a. The resonance effect of the hydroxyl group stabilizes the anionic intermediate
Explanation:
The resonance effect stabilizes the the charge through the delocalization of the pi bonds. The resonance stabilization mainly occurs in the conjugated pi systems.
For example, phenol forms a strong hydrogen bonds than the nonaromatic alcohols as the [tex]$O_2-H_2$[/tex] dipole present in the hydroxyl group is being stabilized by the presence of the aromatic ring of phenol.
Thus the resonance effect of the hydroxyl group stabilizes the anionic intermediate.
A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pressure of the gas after it is heated to 278 K?
Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.
Explanation:
Given: [tex]T_{1}[/tex] = 178 K, [tex]P_{1}[/tex] = 0.3 atm
[tex]T_{2}[/tex] = 278 K, [tex]P_{2}[/tex] = ?
According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.
Formula used to calculate the pressure is as follows.
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\[/tex]
Substitute the values into above formula is as follows.
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm[/tex]
Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.
The blending together of some genes is called:
Answer:
its called molding
Explanation:
What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?
Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
Explanation:
Given: [tex]V_{1}[/tex] = 61 L, [tex]T_{1}[/tex] = 183 K, [tex]P_{1}[/tex] = 0.60 atm
At STP, the value of pressure is 1 atm and temperature is 273.15 K.
Now, formula used to calculate the new volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L[/tex]
Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
A container holds 40.0 mL of nitrogen at 30° C and at a constant pressure.
Find its volume if the temperature increases to 80° C?
Answer:
The correct answer is - 46.60 mL.
Explanation:
To find the volume of the gas at its new increased temperature we need to use Charl Law that shows the direct relationship between Volume and Temperature while Pressure remains constant.
V1 = 40 ml
T1 = 30 degree C + 273 = 303 K
V2 = ?
T2 = 80 degree C + 273 = 353 K
Charl Equation is:
V 1/T 1 = V 2/ T 2
(V1) * (T2)/ T1= V2
placing value:
40*353/303 = V2
= 14120/303
Vf = 46.60 mL
When 155 g of water at 20 ᵒC is mixed with 75 g of water at 80 ᵒC, assuming no heat is
lost to the surrounding what is the final temperature of the water. (Specific heat capacity
of water is 4.18 J/°C·g)
Answer:
39.57 °C
Explanation:
Applying,
Heat gain = Heat lost.
cm'(t₃-t₁) = cm(t₂-t₃)................. Equation 1
Where c = specific heat capacity of water, m = mass of the hotter water, m' = mass of the colder water, t₁ = initial temperature of the colder water, t₂ = initial temperature of the hotter water, t₃ = Final Temperature.
Equation 1 above can futher be simplified to
m'(t₃-t₁) = m(t₂-t₃)................. Equation 2
From the question,
Given: m' = 155 g, m' = 75 g, t₁ = 20 °C, t₂ = 80 °C
Substitute these values into equation 2
155(t₃-20) = 75(80-t₃)
Solve for t₃
155t₃- 3100 = 6000-75t₃
155t₃+75t₃ = 6000+3100
230t₃ = 9100
t₃ = 9100/230
t₃ = 39.57 °C
Assuming no heat is lost to the surrounding, the final temperature of the water is 39.57°C.
Given the following data:
Mass of water 1 = 155 gInitial temperature of water 1 = 20°CMass of water 2 = 75 gInitial temperature of water 2 = 80°CSpecific heat capacity of water = 4.18 J/g°CTo determine the final temperature of the water, assuming no heat is lost to the surrounding:
The quantity of heat lost by the water 1 = The quantity of heat gained by the water 2.
[tex]Q_{lost} = Q_{gained}\\\\m_1c\theta = m_2c\theta\\\\m_1\theta = m_2\theta\\\\m_1(T_2 - T_1) = m_2(T_3 - T_2)[/tex]
Substituting the given parameters into the formula, we have;
[tex]155(T_2 -20)=75(80-T_2 )\\\\155T_2 -3100=6000-75T_2\\\\155T_2 +75T_2=6000+3100\\\\230T_2=9100\\\\T_2=\frac{9100}{230}[/tex]
Final temperature = 39.57°C
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Which molecule will have double bonds?
(4 Points)
Fluorine
Nitrogen
Hydrogen
Oxygen