Answer:
12 pm
Explanation:
Step 1: Given data
Mass of an electron (m): 9.11 × 10⁻³¹ kgAverage speed of an electron (v): 1.9 × 10⁸ m/sSpeed of light (c): 3.00 × 10⁸ m/sPlanck's constant (h): 6.63 × 10⁻³⁴ J.sWavelength of an electron (λ): ?Step 2: Calculate the kinetic energy (E) of an electron
We will use the following expression.
E = 1/2 × m × v²
E = 1/2 × 9.11 × 10⁻³¹ kg × (1.9 × 10⁸ m/s)²
E = 1.6 × 10⁻¹⁴ J
Step 3: Calculate the wavelength of an electron
We will use the following expression.
E = h × c / λ
λ = h × c / E
λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 1.6 × 10⁻¹⁴ J
λ = 1.2 × 10⁻¹¹ m
Step 4: Convert "λ" to picometers
We will use the relationship 1 m = 10¹² pm.
1.2 × 10⁻¹¹ m × (10¹² pm/1 m) = 12 pm
The wavelength of the beam of electrons is 3.8 pm.
Using the De Broglie relation;
λ = h/mv
λ = wavelength
h = Plank's constant
m = mass of the object
v = speed of the object
Now;
λ = ?
h = 6.6 × 10^-34 Js
m = 9.11 × 10^-31 kg
v = 1.9 × 10^8 m/s
So;
λ = 6.6 × 10^-34 Js/ 9.11 × 10^-31 kg × 1.9 × 10^8 m/s
λ = 6.6 × 10^-34 /1.73 × 10^-22
λ = 3.8 × 10^- 12 m or 3.8 pm
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What happens when the elements in group 2 react with iodine
Answer:
Yah
Explanation:
Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6% Write the empirical chemical formula of X.
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Answer:
The empirical formula of X is C3H4O3.
Explanation:
Lamark goes to a bowling alley to test his belief that rolling a bowling ball slowly will result in knocking down more pins. He gently rolls his bowling ball towards the bowling pins lined up at the end of the lane. The bowling ball stops in the middle of the lane without striking the pins. Which statement below best explains what happened?
Answer:
The ball's kinetic energy is transforming to potential energy.
Explanation:
the ball is stopping resulting in potential energy
Someone plz hello me ASAP it would be appreciated on question 2 and 3 btw
Answer:
2 should be potential energy and 3 should be A
Explanation:
How many valence electrons must a lithium atom lose to obtain a complete valence shell?
A. one
B. two
C. three
D. four
Answer:
C. three
Explanation:
Calculate the pH of a 0.46 M solution of NH 4Cl. ( K b for NH 3 = 1.8 × 10 –5)
a.
9.59
b.
4.41
c.
4.80
d.
0.34
e.
9.20
pH of the solution = 4.8
Further explanationNH₄Cl is a salt of a weak base and a strong acid so that it will be partially hydrolyzed
Reaction:
NH₄Cl -----> NH₄⁺ + Cl⁻
NH₄ will be hydrolyzed in water
NH₄⁺ + H₂O ------> NH₃ + H₃O⁺ (H⁺)
This solution releases H⁺ ions so they are acidic
Formula
[tex]\tt [H^+]=\sqrt{\dfrac{Kw}{Kb}.M }[/tex]
Kw = The water equilibrium constant = 10⁻¹⁴
Kb = the ionization constant of a base = 1.8.10⁻⁵
M = 0.46
[tex]\tt [H^+]=\sqrt{\dfrac{10^{-14}}{1.8.10^{-5}}\times 0.46 }[/tex]
[H+]=√2.55 x 10⁻¹⁰ = 1.599 x 10⁻⁵
pH = - log [H+]
pH = 5-log 1.599 = 4.796≈ 4.8
The degree of unsaturation, or index of hydrogen deficiency, is the number of pi bonds plus rings in a molecule. Specify the degree of unsaturation (index of hydrogen deficiency) of the following formulas:
(a) C5H6
(b) C10H6Cl2
(c) C4H3NO
Explanation:
The degree of unsaturation is given as;
Degree of unsaturation = (2C + 2 + N - H - X) / 2
(a) C5H6
Using the formular above;
Degree of unsaturation = [ 2(5) + 2 - 6] / 2
Degree of unsaturation = [ 12 - 6 ] / 2 = 3
(b) C10H6Cl2
Using the formular above;
Degree of unsaturation = [ 2(10) + 2 - 6 -2 ] / 2
Degree of unsaturation = [ 22 - 8 ] / 2 = 7
(c) C4H3NO
Using the formular above;
Degree of unsaturation = [ 2(4) + 2 - 3 -1 ] / 2
Degree of unsaturation = [ 10 - 4 ] / 2 = 3
Note: Oxygen is ignored because its presence has no effect on the degree of unsaturation.
What two types of elements make up an Ionic bond ?
Answer:
Metals and Non-metals
Explanation:
how should the following italicized statement be taken??? elements can be broken further into simpler substances.
a. true
b. false.
Answer:
false
Explanation:
Elements cannot be broken down into a simpler substance.
How many lbs are in 5 kilograms
Answer:
There are approximately 11.0231 pounds in a kilogram
HELP FAST PLZ!!!!! Which phase change allows a substance to transform from a liquid to a
gas?
melting
Ofreezing
O ionization
condensation
deionization
O
evaporation
sublimation
Answer:evaporation
Explanation:
Answer:
Pretty sure its evaporation, if its not I'm very sorry.
Which group beleow is correctly placed in order of least to most electronegative
K, Si, O, F
Si, K, O, F
F, O, SI, K
F, K, O, Si,
Liz and Michelle are testing which ball will bounce the highest. They test 4 different balls. Each ball is dropped five times from the same height. The height of the bounce is measured. At the end of the experiment, the data for each ball was averaged to obtain an average bounce height.
Is this a well designed experiment?
A) No, it does not give any data.
B) No, they should have used more types of balls.
C) Yes, they tested more than one thing at a time.
D) Yes, they averaged the measurements and had multiple trials.
Answer:
D
Explanation:,
In designing an experiment there must be multiple trials. Multiple trials eliminate errors when the results are averaged.
This gives better results than when single measurements are used to determine results of experiments. The result of this experiment can be regarded as being reliable because it was well designed.
Research Hypatia's achievements in the world
of science.
What is she most known for?
Write down three interesting facts about her
life.
Answer:
See explanation
Explanation:
Hypatia is popular for her work in mathematics. She also did some work in the area of astronomy. Her well know work in mathematics is her ideas about conic sections.
She was born the Theon of Alexandria and she was a professional mathematician in her life time.
She was the greatest mathematician of her time and she was telling leader of the Neoplatonist school of philosophy in Alexandria. By so doing, she conquered the culture of sexism in her time.
She was trained by her father in mathematics and eventually replaced him. She was the last major mathematician in the Alexandrian tradition.
For many years, experiments showed that light had wave-like properties. For this reason, scientists described light
as a wave. More recently, experiments produced data that could not be explained by the wave theory of light. The
idea of particle properties was introduced
Why did the theory of light change over time?
O The theory was too old and needed to be updated
New scientists did not agree with the older theory
O The scientific question changed, so the theory had to change.
O Repeated experimentation introduced new discoveries.
Answer:
Repeated experimentation introduced new discoveries.
Explanation:
Science is not a static body of knowledge, scientific theories change as new evidence emerges from repeated experiments.
For many years, the wave theory of light was generally held by almost all scientists.
Further experiments such as photoelectric effect and Compton effect lent credence to the the particle theory of light, hence the answer
Answer:
D
Explanation:
Repeated experimentation introduced new discoveries.
g A 0.4395 g sample of aluminum reacts according to our experiment to produce alum. 5.1629 g of dried alum crystals are recovered. What is the percent yield of the experiment?
Answer:
92.75%
Explanation:
The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:
[tex]\mathtt{2Al + 2KOH + 4H_2SO_4 +2H_2O \to 2KAl(SO_4)_2 2H_2O +3H_2}[/tex]
From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.
Therefore, the theoretical yield can be determined as:
[tex]=0.4395 \ g Al \times \dfrac{1 \ mol \ Al}{27 \ g Al }\times \dfrac{2 \ mol \ KAl(SO_4)_2}{2 \ mol \ Al}\times \dfrac{294.23 \ KAl(SO_4)_2}{1 \ mol \ KAl(SO_4)_2}[/tex]
= 4.789g of [tex]KAl(SO_4)_2[/tex]
To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.
∴
percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100
percent yield = ( 4.789g / 5.1629g ) × 100%
percent yield = 0.9275 × 100%
percent yield = 92.75%
Thus, the percent yield of the experiment 92.75%
What is the density of water if you have 50.0 grams of water and a volume of 50.0 millimeters
Answer:
[tex]\boxed {\tt 1.0 \ g/mL}[/tex]
Explanation:
Density can be found by dividing the mass by the volume.
[tex]d=\frac{m}{v}[/tex]
The mass of the water is 50.0 grams.
The volume of the water is 50.0 milliliters.
[tex]m= 50.0\ g \\v=50.0 \ mL[/tex]
Substitute the values into the formula.
[tex]d=\frac{50.0 \ g}{50.0 \ mL}[/tex]
Divide.
[tex]d= 1.0 \ g/mL[/tex]
The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³
Is Kool-Aid dissolved in water in element compound or mixture?
Answer:
mixture
Explanation:
Kool-Aid is a mixture of sugar, water, dye, and flavoring.
When 2.16g of H2 reacts with excess O2 by the following equation, 258 kJ of heat are released. What is the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas? 2H2+O2⟶2H2O
Answer:
-241 kJ/mol
Explanation:
Let's consider the reaction of hydrogen with excess oxygen to form water.
2 H₂ + O₂ ⟶ 2 H₂O
When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:
ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol
Green light has a wavelength of 510 nm. What is the wavelength in meters?
5. What would be the temperature change if 12.5 g of water absorbed 35 J of heat?
Explanation:
its 399929939932929292
Find ΔHrxn for the following reaction:
2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)
Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
The standard enthalpy of the reaction is -827.5 kJ/mol
The standard enthalpy of reaction [tex]\mathbf{\Delta H^0_{rxn}}[/tex] is the enthalpy change that happens in a system whenever one mole of the matter is converted through a chemical process under normal conditions.
The given reaction can be expressed as:
2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)
The standard enthalpy can be represented by the equation:
[tex]\mathbf{\Delta H^0_{rxn} = \sum \Delta _f ^0(products) - \sum \Delta _f^0(reactants)}[/tex]
At standard conditions, the standard enthalpies of formation of the given species are:
ΔH°f(PbO(s)) = 219 kJ/molΔH°f(SO₂(g)) = -296.83 kJ/molΔH°f(PbS(s)) = 100.4 kJ/molΔH°f(O₂(g)) = 0 kJ/mol
∴
[tex]\mathbf{ \Delta H^0_{rxn} = \Big[2 mol \times \Delta H^0_f(PbO(s)) + 2 mol \times \Delta H^0_ f(SO_2(g) )\Big] - \Big[2 mol \times \Delta H^0_f (PbS(s))} + \mathbf{ 3 mol \times \Delta H^0_f(O_2(g) )\Big] }}[/tex]
[tex]\mathbf{\Delta H^0rxn = [2 mol \times (-217.32 kJ/mol) + 2 mol \times (-296.83)] - [2 mol \times (-100.4)} \\ \mathbf{+ 3 mol \times 0 kJ/mol]}}[/tex]
[tex]\mathbf{\Delta H^0rxn = -827.5 \ kJ/mol}}[/tex]
Therefore, we can conclude that the standard enthalpy of the reaction is -827.5 kJ/mol
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Which step of mitosis involves the nucleus spitting and nuclear membranes forming around each new nucleus??
Answer:
telophase
Explanation:
2. Determine the possible traits of the calves of : Da red (RR) bull is mated with a red (RR) cow 1 a red * (RR) bullis mated with a white (WW) Cow 2 Da roan * (RW) is mated with a red(RR)Cow 3 3. Illustrate your answers using a Punnett square. 4. Write your answers on the paper.
Describe three factors that could limit the growth of the prairie dog population
Answer:
Changes in environment, food source changes and disease?
Explanation:
I dont know what the answer choices are
Answer:
changes in the environment
Explanation:
like digging big holes.
Hope this helps!
The escape velocity from Earth’s surface is 1.12*10^4 meters per second. At this speed, how many kilometers would a rocket travel in 3 minutes
Answer:
2016 Km.
Explanation:
The following data were obtained from the question:
Speed (S) = 1.12×10⁴ m/s
Time (t) = 3 mins
Distance (d) =.?
Next, we shall convert 3 mins to seconds. This can be obtained as follow:
1 min = 60 s
Therefore,
3 mins = 3 min × 60 s / 1 min
3 mins = 180 s
Next, we shall determine the distance travelled by the rocket. This is illustrated below:
Speed (S) = 1.12×10⁴ m/s
Time (t) = 180 s
Distance (d) =.?
Speed (S) = distance (d) /time (t)
S = d/t
1.12×10⁴ = d/ 180
Cross multiply
d = 1.12×10⁴ × 180
d = 2016000 m
Finally, we shall convert 2016000 m to Km. This can be obtained as shown below:
1000 m = 1 Km
Therefore,
2016000 m = 2016000 m × 1 Km / 1000 m
2016000 m = 2016 Km
Therefore, the rocket will travel 2016 Km in 3 mins
What does not represent a compound
Answer:
Every combination of atoms is a molecule. A compound is a molecule made of atoms from different elements. All compounds are molecules, but not all molecules are compounds. Hydrogen gas (H2) is a molecule, but not a compound because it is made of only one element.Explanation:
A Cu/Cu2 concentration cell has a voltage of 0.22 V at 25 o C. The concentration of Cu2 in one of the half-cells is 1.5 x 10-3 M. What is the concentration of Cu2 in the other half-cell
Answer:
The concentration is [tex] [Cu^{2+}]_a = 10^{-10.269} [/tex]
Explanation:
From the question we are told that
The voltage of the cell is [tex]E = 0.22 \ V[/tex]
Generally the reaction at the cathode is
[tex]Cu^{2+} _{(aq)} + 2e^{-} \to Cu_{s}[/tex] the half cell voltage is V_c = 0.337 V
Generally the reaction at the anode is
[tex]Cu _{(s)} \to Cu^{2+} _{(aq)} + 2e^{-}[/tex] the half cell voltage is V_a = -0.337 V
Gnerally the reaction of the cell is
[tex]Cu_{(s)} + Cu^{2+} _{(aq)} \to Cu^{2+}_{(aq)} + Cu_{(s)}[/tex]
At initial the voltage is V = 0 V
Generally the voltage of the cell at 25°C is
[tex]E = V - \frac{0.0591}{n} log \frac{[Cu^{2+}] _a}{[Cu^{2+}]_c}[/tex]
Here n is number of of electron and it is 2
So from the question we are told that one cell has a concentration 1.5 x 10-3 M
Let assume it is [tex][Cu^{2+}]_c[/tex]
So
[tex]0.22= 0 - \frac{0.0591}{2} log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }[/tex]
=> [tex]-7.445 = log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }[/tex]
=> [tex]-7.445 = log [Cu^{2+}_a] - log [1.5*10^{-3}][/tex]
=> [tex]-7.445 + log [1.5*10^{-3} = log [Cu^{2+}_a] [/tex]
=> [tex]-7.445 - 2.824 = log [Cu^{2+}_a] [/tex]
Taking the antilog
=> [tex] [Cu^{2+}]_a = 10^{-10.269} [/tex]
=> [tex] [Cu^{2+}]_a = 5.38 *10^{-11} \ M [/tex]
A 5-kg quantity of radioactive isotope decays to 2 kg after 10 years. Find the decay constant of the isotope.k = _____
Answer:
k = -0.09165 years^(-1)
Explanation:
The exponential decay model of a radioactive isotope is generally given as;
A(t) = A_o(e^(kt))
Where;
A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years
We are given;
A_o = 5 kg
A(10) = 2kg
t = 10 years
Thus;
A(10) = 2 = 5(e^(10k))
Thus;
2 = 5(e^(10k))
2/5 = (e^(10k))
0.4 = (e^(10k))
In 0.4 = 10k
-0.9164 = 10k
k = -0.9164/10
k = -0.09165 years^(-1)
1) Why does polysubstitution frequently occur with Friedel-Crafts alkylation but not with Friedel-Crafts acylation?2) If Benzene is reacted with t-butyl chloride, why is the chief disubstituted product the para rather than ortho isomer?
Answer:
See explanation
Explanation:
Friedel-Crafts alkylation frequently undergo polysubstitution because the process leads to the addition of an electron donating alkyl group, which activates the benzene ring to further alkylation. The acyl group rather deactivates the ring towards electrophilic substitution.
When benzene is reacted with t-butyl chloride, the chief disubstituted product is the para rather than ortho isomer because the increased bulk of the tert-butyl group hinders attack at the ortho-sites.
