42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.
The most popular scales include the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of degrees Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes. One of the United Nations System of Units' (SI) seven base units is the kelvin.
Q=725 J
m=55.0 g
c=0.900 J/(°C⋅g)
ΔT=final temperature - initial temperature
ΔT=(x−27.5)°C
725 J=55.0 g⋅0.900 J/(°C⋅g)(x−27.5)
X=42.1°C
Therefore, 42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.
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3.01 × 1023 molecules H2O
Answer:
0.5 mole
Explanation:
The question isn't even clear
But I'm guessing you want to ask the number of moles
n= Number of molecules/ Avogadros number
n= 1/2
A 11.0 mLmL sample of 0.30 MHBrMHBr solution is titrated with 0.16 MNaOHMNaOH. Part A What volume of NaOHNaOH is required to reach the equivalence point? Express the volume to two significant figures and include the appropriate units. nothingnothing
Answer:
21 mL of NaOH is required.
Explanation:
Balanced reaction: [tex]HBr+NaOH\rightarrow NaBr+H_{2}O[/tex]
Number of moles of HBr in 11.0 mL of 0.30 M HBr solution
= [tex](\frac{0.30}{1000}\times 11.0)[/tex] moles = 0.0033 moles
Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.
So, number of moles of NaOH in V mL of 0.16 M NaOH solution
= [tex](\frac{0.16}{1000}\times V)[/tex] moles = 0.00016V moles
According to balanced equation-
1 mol of HBr is neutralized by 1 mol of NaOH
So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH
Hence, [tex]0.00016V=0.0033[/tex]
[tex]\Rightarrow V=\frac{0.0033}{0.00016}=21[/tex]
So, 21 mL of NaOH is required.
Enter your answer in the provided box. A sample of an unknown gas effuses in 14.5 min. An equal volume of H2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass = 71.76 g/mol
Explanation:
The relationship between molar mass and rate of effusion is given as;
Vh / Vu = √ (Mu / Mh)
The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.
Rate = volume / time (Assuming Volume = 1)
Vh = Rate of effusion of Hydrogen = 1 / 2.42
Vu = Rate of effusion of unknown gas = 1 / 14.5
Mh = Molar mass of hydrogen = 2
Mu = Molar mass of unknown gas = x
Substituting into the formular, we have;
(1 / 2.42) / (1 / 14.5) = √ ( x / 2)
5.99 = √ ( x / 2)
35.88 = x / 2
x = 71.76
A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4
Answer:
N2H4
Explanation:
A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.
Now consider the hydride N2H4.
N2H4(g) -----> N2(g) + 2H2(g)
The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.
The molecular mass of the compound is;
N2H4= 2(14) + 4(1)= 28+4= 32
Since
molecular mass= 2 vapour density
Vapour density= molecular mass/2
Vapour density= 32/2
Vapour density = 16
Therefore the hydride of nitrogen referred to in the question is N2H4
Calculate the percent saturated fat in the total fat in butter
Indicate whether each of the following indicates that a physical or chemical change has taken place when a piece of magnesium metal is studied: (a) Can be cut into tiny pieces (b) Fizzling occurs when placed water (c) Light is emitted when burned (d) Turns to ash
Answer:
a) Can be cut into tiny pieces - Physical Change
b) Fizzling occurs when placed water -Chemical Change
c) Light is emitted when burned -Chemical Change
d) Turns to ash -Chemical Change
Explanation:
CHEMISTRY HELP!
using only the periodic table, determine the charge on the ion that is formed by arsenic.
The ion charge is:
a. -3
b. -2
c. -1
d. 0
e. +1
f. +2
g. +3
also what is it for elements lithium and strontium?
Answer:
A
Explanation:
Arsenic is in the same group as Nitrogen - group 5. They all have 5 valence electrons in their outermost shell. To achieve its most stable state - 8 valence electrons (octet rule - elements are most stable when the entire shell is filled) - arsenic needs to gain 3 electrons. Since electrons have a negative charge, the charge of an As ion would be -3.
Try observing the periodic table and how many valence electrons that each element has. From there, you can determine the charges of the elements lithium and strontium. You can guess, I'll help you with those once you attempt to find the charge of those ions.
A 1.00 liter solution contains 0.42 moles nitrous acid and 0.32 moles sodium nitrite .
If 0.16 moles of nitric acid are added to this system, indicate whether the following statements are true or false.
(Assume that the volume does not change upon the addition of nitric acid.)
A. The number of moles of HNO2 will decrease.
B. The number of moles of NO2- will remain the same.
C. The equilibrium concentration of H3O+ will increase.
D. The pH will decrease.
E. The ratio of [HNO2] / [NO2-] will increase
Answer:
E. The ratio of [HNO2] / [NO2-] will increase
D. The pH will decrease.
Explanation:
Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .
pH = pka + log [ salt] / [ Acid ]
= 3.4 + log .32 / .42
= 3.4 - .118
= 3.282 .
Now .16 moles of nitric acid is added which will react with salt to form acid
HNO₃ + NaNO₂ = HNO₂ + NaNO₃
concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )
concentration of nitrous acid = .42 + .16 = .58 M
concentration of salt = .32 - .16 = .16 M
ratio of [HNO₂ ] / NO₂⁻]
= .42 / .32 = 1.3125
ratio of [HNO₂ ] / NO₂⁻] after reaction
= .42 + .16 / .32 - .16
= 58 / 16
= 3.625 .
ratio will increase.
Option E is the answer .
pH after reaction
= 3.4 + log .16 / .58
= 2.84
pH will decrease.
Best example of potential energy?
Answer:
water stored in a dam
Explanation:
when the water is in dam it is ready to move bit is not moving
How is excitation in spectroscopy brought about
Answer: the exciation of molecules is brount by absorption of energy in spectroscpy
Explanation:
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:
2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)
Suppose the cell is prepared with 1.87 M MnO−4 and 1.37 M H+ in one half-cell and 3.23 M Mn+2 and 6.62 M Pb+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Answer:
1.63 V
Explanation:
Let us state the reaction equation again for the purpose of clarity;
2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)
The reduction potentials for the two half reaction equations are;
MnO 4 - (aq) + 8H + (aq) + 5e - → Mn2+(aq) + 4H2O(l) Eo=1.51 V
Pb2+(aq) + 2e - → Pb(s) Eo= -0.13 V
E°cell = E°red – E°Ox
E°cell = 1.51 - (-0.13)
E°cell = 1.51 + 0.13
E°cell = 1.64 V
But Q= [Mn^2+]^2 [Pb^2+]^5/[MnO4^-]^2 [H^+]^16
Q= [3.23]^2 [6.62]^5/[1.87]^2 [1.37]^16
Q= 10.43 × 12714.22/3.4969 × 154
Q= 132609.3/538.5226
Q= 246.25
From Nernst equation
E= E° - 0.0592/n log Q
Where n=10
E= 1.64- 0.0592/10 log 246.25
E= 1.64-0.0142
E= 1.63 V
n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a phenolphthalein end point. Calculate the molarity of the acetic acid solution, and the percent (by weight) of acetic acid in the solution (assuming its density to be 1.00 g/mL).
Answer:
Molarity: 0.111M
% (w/w): 0.666
Explanation:
The reaction of NaOH with acetic acid (CH₃COOH) is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.
As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:
0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:
4.169x10⁻³ moles of CH₃COOH.
Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:
4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M
As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:
4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid
Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.
Thus, percent by weight is:
0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)
The molarity of acetic acid is 0.11M and the percent by weight is 0.666%.
How we calculate molarity?Molarity of any solution is used to define their concentration and it will be calculated as:
M = n/V, where
n = moles
V = volume
Molarity of acetic acid will be calculated as:
M₁V₁ = M₂V₂, where
M₁ = molarity of acetic acid = ?
V₁ = volume of acetic acid = 37.54mL = 0.037L
M₂ = molarity of NaOH = 0.1062M
V₂ = volume of NaOH = 39.26mL = 0.039L
On putting all these values on the above equation we can calculate the molarity as:
M₁ = (0.1062)(39.26) / (37.54) = 0.11M
Now we calculate the moles of acetic acid by using the molarity formula as:
n = 0.11M × 0.037L = 0.00407 moles
Molar mass of acetic acid = 60g/mole
Mass of 0.00407 moles of acetic acid = 4.1x10⁻³ moles×(60g / mol) = 0.2501 g
Density of solution = 1.00 g/mL
So, 37.54mL in 1g/mL = 37.54g/mL
Percent by weight will be calculated as:
%w/w = 0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)
Hence, molarity and %(w/w) of acetic acid is 0.11M and 0.666% respectively.
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A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.
Answer:
1.The electrode on the right is positive
2. 0.058V
Explanation:
The above cell is a concentration cell.
A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).
Part 2: Please, see the attachment below for the calculations.
A chemistry student weighs out 0.306 g of citric acid (H3C6H5O7) , a triprotic acid, into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits
Answer:
THE VOLUME OF NaOH NEEDED TO BE ADDED TO CITRIC ACID TO REACH THE EQUIVALENT POINT IS 4.725 L
Explanation:
The titration is between citric acid (H3C6H507) and NaOH
mass of citric acid = 0.306 g
Volume of citric acid = 250 mL = 250 /1000 = 0.25 L
Concentration of NaOH = 0.1000 M
Volume = unknown
First calculate the molar mass of citric acid
( 1 * 3 + 12* 6 + 1*5 + 16*7) = (4 + 72 + 5 + 112) = 193 g/mol
Since,
Concentration in moles/dm3 = concentration in g/dm3 / RMM
So the molarity of citric acid is:
Molarity = 0.306g / 0.25dm3 / Rmm
Molarity = 1.224g/dm3 / 193 g/mol
Molarity = 0.0063 M
Equation for the reaction is:
C3H5O(COOH)3 + 3NaOH → Na3C3H5O(COO)3 + 3H2O
Using the formula:
CaVa / CbVb = na/ nb
Ca = 0.0063 M
Cb = 0.1000 M
Va = 0.25 L
Vb = unknown
na = 1
nb = 3
Vb = Ca Va nb/ Cb na
Vb = 0.0063 * 0.25 * 3 / 0.1000 * 1
Vb = 0.4725 / 0.1000
Vb = 4.725 L
The volume of NaOH needed to reach the equivalent point is therefore 4.725 L
Consider the reaction C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (l) in which 10.8 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/oC (including its water). The temperature inside the calorimeter was found to increase by 20.5 oC. Based on this information, what is the heat of this reaction per mole of sucrose? Enter your answer numerically, in terms of kJ/
Answer:
THE HEAT OF THIS REACTION PER MOLE OF SUCROSE IS 4868.75 KJ OF HEAT.
Explanation:
To answer this question:
First calculate the total heat given off by sucrose:
Total energy/ heat = heat capacity * change in temperature
Heat capacity = 7.50 kJ/ °C
Change in temperature = 20.5 °C
Heat = 7.50 kJ * 20.5 °C
Heat = 153.75 kJ of heat.
Next is to calculate the heat of reaction per mole of the sucrose
Equation of the reaction:
C12H22011 (s) + 12 O2 (g) ---------> 12 CO2 (g) + 11 H20(l)
Since 1 mole of sucrose will be the molar mass of sucrose, then we should calculate the molar mass of sucrose.
Molar mass of sucrose = ( 12* 12 + 1 * 22+ 16*11) g/mol
Molar mass = 342 g/mol of sucrose
Since 10.8 g of sucrose produces 153.75 kJ of heat, 342 g will produces how many joules of heat?
10.8 g of sucrose = 153.75 kJ of heat
342 g of sucrose = ( 342 * 153.75 kJ / 10.8)
= 52 582.5 kJ / 10.8
= 4868.75 kJ of heat
So therefore, 1 mole of sucrose will produce 4868.75 kJ of heat.
Give the full electron configuration for sulfur.
electron configuration:
Answer:
[Ne] 3s² 3p⁴
Electrons per shell: 2,8,6
Answer : 2,8,6 (Sulphur Atom)
2,8,8 (Sulphur Ion)
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 500 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg
Answer:
Explanation:
a )
m = m₀ [tex]e^{-\lambda t[/tex]
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ [tex]e^{- 0.0004358 t}[/tex]
b )
m = 50 x [tex]e^{-.0004358\times 500}[/tex]
= 40.21 mg .
c )
40 = 50 [tex]e^{-.0004358t[/tex]
.8 = [tex]e^{-.0004358t[/tex]
[tex]e^{.0004358t[/tex] = 1.25
.0004358 t = .22314
t = 512 years .
• Briefly discuss the cause of errors in the measurements
A quantity of 2.00 × 102 mL of 0.662 M HCl is mixed with 2.00 × 102 mL of 0.331 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 22.00°C. For the process below, the heat of neutralization is −56.2 kJ/mol. What is the final temperature of the mixed solutions? H+(aq) + OH−(aq) → H2O(l)
Answer:
Final temperature of the solution = 26.43°C
Explanation:
Concentration of HCl = 0.662 M, Volume = 200 mL= 0.200 L
Concentration of Ba(OH)₂ = 0.331 M, Volume = 200 mL = 0.200 L
Initial temperature of solution = 22.00°C
Specific Heat capacity of water = 4.184 J/g°C
Heat of neutralization = -56.3 KJ/mol of H₂O produced.
The full calculations is found in the attachment below
1. Which statement describes the particles of an ideal gas, based on the
kinetic molecular theory?*
O There are attractive forces between the particles.
O The particles move in circular paths.
O The collisions between the particles reduce the total energy of the gas.
О
The volume of the gas particles is negligible compared with the total volume of the
gas.
Answer:the volume of the gas particles is negligible compared with the total volume of the gas.--D
Explanation:
According to the Kinetic Molecular Theory for ideal gases, it states that
--Gases are composed of larges molecules which are in constant random motion in a straight line
--The volume of the gas particles is negligible compared to the total volume in which the gas is contained.
-----The Attractive and repulsive forces between gas molecules is insignificant ie There are no interactive forces.
----The collisions of the particles are perfectly elastic and energyis being transferred between the particles but the total energy remaining constant
From the statements of the kinetic Molecular theory of ideal gases, it can be seen that the statement which describes the particles of an ideal gas is option D which is The volume of the gas particles is negligible compared with the total volume of the gas--- ---This gives the reason why gases can be compressed. Since there are no inter molecular forces between them. The particles of an ideal gas will move at the same random motion resulting to high pressures, compressing the gas and making the volume negligible or insignificant.
RUIGA GIRLS
CHEMISTRY FORM 3. 23/06/2020
MR. GICHURU
IZ
1
Narne the elements present in
Common salt
(2 miks)
Hydrated copper (11) Sulphate.
(2 ks)
Sulphuric (VI) acid,
2 Why is a reaction between zinc metal and Nitric acid not suitable for preparing
hydrogen gae in the laboratory
(2 mi)
(1 m)
3.
What is relative atomic mass?
b)
Define 'isotopes
c)Determine the relative atomic mass of element K whose isotople misure occur in
the proportione:
(2 marks)
What is the probability that an offspring will have a
heterozygous genotype? |
Answer:
25,50,25
Explanation:
A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution
Answer:
[tex]0.745~M[/tex]
Explanation:
In this case, we have a dilution problem. So, we have to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
Now, we have to identify the variables:
[tex]C_1~=~14.9~M[/tex]
[tex]V_1~=~25~mL[/tex]
[tex]C_2~=~?[/tex]
[tex]V_2~=~0.5~L[/tex]
Now, we have different units for the volume, so we have to do the conversion:
[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]
Now we can plug the values into the equation:
[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]
I hope it helps!
A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C
Answer:
FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL
Explanation:
From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:
k = Ae^ -Ea/RT
At initial temperature T1, the initial rate constant is (k1)
At final temperature T2, the final rate constant is k2
For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.
That is, k2 / k1 = 2 (rate is doubled)
Equating this into the Arrhenius equation, we have:
k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)
2 = e^ (-Ea / R) (1 / T2 = 1 / T1)
Taking the natural logarithm of both sides:
ln 2 = - (Ea / R) (1 / T2 - 1 / T1)
Making Ea the subject of the formula, we obtain:
Ea = - (ln 2 R / (1 / T2- 1 / T1))
Let T1 = 25 C = 25 + 273 K = 298 K
T2 = 35 C = 35 + 273 K = 308 K
R = 8.314
So,
Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))
Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)
Ea = - 5.7616 / -0.00011
Ea = 52 378,18 J / mol
So therefore, the activation energy Ea is 52.4 kJ/mol.
What is the number of valence electrons in a nitrogen atom in the ground state
Answer: 5
Explanation:
It just is
Answer:
5
Explanation:
Bc valence electron means last # in the electron configuration
.Draw the born-Haber lattice energy cycle for sodium chloride. Explain the concept of resonance using the nitrate ion structure.
Answer:
Born-Haber cycle is consist on four to five steps. 1: ionization energy 2: electron affinity 3: dissociation energy 4: sublimation energy and last is Hess law.Nitrate ion have 3 localized sigma bonds and 1 delocalized pie bond according to the resonance structure.Explanation:
Step 1: NaCl(s) → Na(s) + 1/2 Cl2(g) ΔHf (ionization energy) in this step energy is required to change the phase of the compound
Step 2: Na(s) + 1/2 Cl2(g) → Na(g) + 1/2 Cl2(g) ΔHa (elements needed to be in gaseous state for born-haber cycle so metal changes from solid to gas state by changing the enthalpy.
Step 3: Na(g) + 1/2 Cl2(g) → Na(g) + Cl (g) 1/2ΔHd
Step 4: Na(g) + Cl(g) → Na⁺(g) + Cl⁻(g) IE+EA ( in this step both ionization energy and electron affinity was involved because in metal (Na) electron is added which needs the energy and this energy draw from the step 3 and Chlorine require releasing electron to be in ionic state so when electron leaves the orbit energy releases.
Step 5: final step is Hess Law which is the combination of all the steps which step 4 again go back to step 5 and this cycle continues by repeating same steps Na⁺(g) + Cl⁻(g)→NaCl(s)
at this step heat of formation is calculated
Heat of formation= atomization energy+ dissociation energy+ sum of ionization energies + sum of electron affinity + lattice energy.
2: if we look at the electron configuration of the nitrogen it has 5 electrons in its outermost shell which indicates it can make 5 bonds 4 bonds and 1 lone pair usually and Oxygen has 6 electrons in its outermost shell. So nitrate ion have the total number of 24 electrons including the 1 electron which shows on the compound.
So when they make nitrate ion NO₃⁻¹ it shows that nitrate has 3 resonance structures. Nitrogen's three sigma bonds are attached to oxygen and fourth one make 1 pie bond which can rotate, delocalized and change its position anytime from one Oxygen atom to other oxygen atom.
The metalloid that has three valence electrons is .
Answer:
Boron
Explanation:
Answer: a. boron
Explanation: bc the Boron Family is named after the first element in the family. atoms in this family have 3 valence electrons. this family includes a metalloid (boron), and the rest are metals.
The compound ClF contains Group of answer choices polar covalent bonds with partial negative charges on the Cl atoms. ionic bonds. nonpolar covalent bonds. polar covalent bonds with partial negative charges on the F atoms.
Answer:
polar covalent bonds with partial negative charges on the F atoms.
Explanation:
A covalent bond could be polar or nonpolar depending on the relative electro negativity difference between the two bonding atoms. In this case, the bonding atoms are chlorine and fluorine.
In the Pauling's scale, fluorine has an electro negativity value of 3.98 while chlorine has an electro negativity value of 3.16. The difference in electro negativity between the two atoms is about 0.82. This magnitude of electro negativity difference between the two bonding atoms correspond to the existence of a polar covalent bond in the molecule.
The direction of the dipole depends on the relative electro negativity values of the two bonding atoms. Since fluorine is more electronegative than chlorine, the fluorine atom will be partially negative and the chlorine atom will be partially positive accordingly.
The compound ClF (chlorine monofluoride) contains polar covalent bonds with partial negative charges on the F atoms. Therefore, option D is correct.
In ClF, chlorine (Cl) is more electronegative than fluorine. As a result, the shared electrons in the Cl-F bond are pulled closer to the chlorine atom, creating a partial negative charge on the fluorine atoms and a partial positive charge on the chlorine atom.
This polarity in the Cl-F bond gives the molecule an overall polarity, making it a polar molecule. Thank you for pointing out the error, and I apologize for any confusion caused.
Thus, option D is correct.
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Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium tertbutoxide in tertbutanol.
Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium ethoxide in ethanol.
Answer:
see explanation below
Explanation:
In the first case, we have a reaction where we have the 3-chloro-3-methylpentane reacting with t-butoxide. The t-butoxide is a very voluminous base, so the strength of substracting a hydrogen atom is reduced. Therefore, the reaction taking place here will be an E2 but instead of substracting the hydrogen from the carbons 2 or 4, it will substract it from the methyl group, cause it has less steric hindrance there and the reaction will go faster.
In the second case, the sodium ethoxide is a strong base, so it will rapidly substract an atom of hydrogen from carbon 2 or 4 to form the (Z) - 3 - methyl - 2- pentene and the substitution product.
Look picture for mechanism and products.
Chemical formula for copper gluconate I have 1.4g of Copper gluconate. There is .2g of copper within the copper gluconate. Determine the chemical formula for Copper gluconate with the given information: Copper Gluconate: Cu(C6H11O?)? Cu = 63.55 g/mol H = 12.01 g/mol O = 1.008 g/mol Cu = 63.55 g/mol
Answer:
The simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
Explanation:
Given mass of sample = 1.4 g
mass of copper in the sample = 0.2 g
mass of the gluconate =1.4 - 0.2 = 1.2 g
The mole ratio is determined first using the formula;
mole ratio = reacting mass / atomic mass
atomic mass of copper = 63.55
mass of gluconate, C₆H₁₁O₇ = 12*6 + 1*11 + 16*7 = 195 g/mol
mole ratio ( copper : gluconate) = 0.2/63.55 : 1.4/195
mole ratio ( copper : gluconate) = 0.003 : 0.007
convert to whole number ratios by dividing with the smallest ratio
mole ratio ( copper : gluconate) = 0.003/0.003 : 0.007/0.003
mole ratio ( copper : gluconate) = 1 : 2
Therefore, the simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂