Answer:
[tex]\lambda=3451*10^{10}m[/tex]
Explanation:
From the question we are told that:
Energy state [tex]e=3.50 eV[/tex]
Time [tex]t=2ms[/tex]
Generally the equation for energy of Photon is mathematically given by
[tex]E=e-e_0[/tex]
[tex]E=3.6*10^{-19}J[/tex]
[tex]E=5.7*10^{-19}J[/tex]
Generally the equation for Wave-length of Photon is mathematically given by
[tex]\lambda=\frac{hc}{E}[/tex]
[tex]\lambda=\frac{6.626*10^{-34}*3*10^8}{5.76*10^{-19}}[/tex]
[tex]\lambda=3451*10^{10}m[/tex]
What Volume of silver metal will weigh exactly 2500.0g. The density of silver
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3
I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even though solubility rules would not lead one to predict so because potassium nitrate is obviously soluble and so should copper (II) iodide. One can deduce from the formation of a precipitate that copper is reduced. Write a proposed reaction for the oxidation reduction of copper (II) iodide. Justify the choice of the substance that reduces the copper based on experimental evidence. Also, justify the choice using the atomic structure of potassium ion and iodide ion.
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
What is the mole of 98 mL of carbon dioxide gas at 36°C and 795 torr?
R = 0.0821 Latm/molk
Round to the thousandth place.
The following reaction is not an oxidation-reduction reaction: Fe(s) + 2Hl(aq) --- Fel (aq) + H_(8) Select one: O True O False
Explanation:
the reaction is indeed an oxidation reduction reaction
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
a Leaning towards the speaker
Which combination of reagents used in the indicated order with benzene will give m-nitropropylbenzene? a. 1) CH3CH2CH2Cl/AlCl3, 2) HNO3/H2SO4 b. 1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3 c. 1) CH3CH2CH2Cl/AlCl3, 2) HNO3/H2SO4, 3) Zn/HCl d. 1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3, 3) Zn/HCl
Answer:
1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3
Explanation:
Benzene is a stable aromatic compound hence it undergoes substitution rather than addition reaction.
When benzene undergoes substitution reaction, the substituent introduced into the ring determines the position of the incoming electrophile.
If I want to synthesize m-nitropropylbenzene, I will first carry out the nitration of benzene using HNO3/H2SO4 since the -nitro group is a meta director. This is now followed by Friedel Craft's alkykation using CH3CH2CH2Cl/AlCl3.
1)Consider the reactions observed when NaOH(aq) and NH3(aq) were added to the solutions containing Zn2 (aq): a)What is the identity of the precipitate formed when the NaOH(aq) and NH3(aq) were added dropwise (limited amounts added)
Answer:
A white gelatinous precipitate is observed in each case.
Explanation:
Qualitative analysis in chemistry is mostly used to identify the ions present in a sample by adding certain reagents. The observation after adding the reagent often leads to an inference.
When NaOH is added to a solution containing Zn^2+ in drops, a white gelatinous precipitate is observed.
When NH3(aq) is added in drops to a solution containing Zn^2+, a white gelatinous precipitate is also observed.
Identify what reagents you would use to achieve each transformation: Conversion of 2-methyl-2-butene into a secondary alkyl halide. Br2, ROOR Br2, H2O HBr, ROOR HBr Conversion of 2-methyl-2-butene into a tertiary alkyl halide. Br2, H2O HBr Br2, ROOR HBr, ROOR
Answer:
Conversion of 2-methyl-2-butene into a secondary alkyl halide - ROOR, HBr
Conversion of 2-methyl-2-butene into a tertiary alkyl halide - HBr
Explanation:
The addition of HBr to 2-methyl-2-butene occurs in accordance to Markovnikov rule in the absence of peroxide.
According to Markovnikov rule; ''the negative part of the addendum is attached to the carbon atom bearing the least number of hydrogen atoms.'' Following the Markovnikov rule, the tertiary alkyl halide is obtained.
In the presence of peroxide, this rule is not followed and the reaction proceeds in an anti-Markovnikov way to yield a secondary alkyl halide.
Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 hours for an adolescent. What fraction of the original reactant concentration will remain after 8 hours if the original concentration was 8.4 x 10-5 M.
Explanation:
The given data is:
The half-life of gentamicin is 1.5 hrs.
The reaction follows first-order kinetics.
The initial concentration of the reactants is 8.4 x 10-5 M.
The concentration of reactant after 8 hrs can be calculated as shown below:
The formula of the half-life of the first-order reaction is:
[tex]k=\frac{0.693}{t_1_/_2}[/tex]
Where k = rate constant
t1/2=half-life
So, the rate constant k value is:
[tex]k=\frac{0.693}{1.5 hrs}[/tex]
The expression for the rate constant is :
[tex]k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}[/tex]
Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:
[tex]\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6[/tex]
Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
Which of the following statements is/are correct? 1. There are 5 orbitals in
the d subshell. 2. The d orbitals can accommodate 14 electrons. 3. The first
shell contains s and p orbitals. 4. The s orbital can accommodate 2 electrons.
A. 1 and 4
B. 2 and 3
C. 3 and 4
D. 2 and 4
If 1 mol of ferric oxide reacts with 3 moles of carbon monoxide to yield 2 mols of iron and 3 mols of carbon dioxide, how much CO will be needed to completely react with 50.26 g of ferric oxide?
Answer:
26.4g
Explanation:
The balanced chemical equation as stated in this question is given as follows:
Fe2O3 + 3CO → 2Fe + 3CO2
According to this balanced equation, 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).
We need to change 50.26 g of ferric oxide to moles by using the formula;
mole = mass/molar mass
Molar mass of Fe2O3 = 56(2) + 16(3)
= 112 + 48
= 160g/mol
mole = 50.26/160
mole = 0.314mol of Fe2O3
If 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).
Hence, 0.314 mol of Fe2O3 will completely react with (0.314 × 3) mol of CO
0.314 × 3 = 0.94 mol of CO
molar mass of CO = 12 + 16 = 28g/mol
mole = mass/molar mass
mass = mole × M.M
mass = 0.94 × 28
mass = 26.4g of CO
In the graphic, 195 represents the _______.
195 Pt
78
A. Atomic Mass
B. Atomic Number
C. Neutron Number
Answer:
ITS ANSWER IS
OPTION B. ATOMIC NUMBER
HI HAVE A NICE DAY
At constant temperature and pressure, if 0.4 mole of a gas. A occupies 220 ml, and x mole of B gas occupies 120 ml. what is the number of moles of gas B?
Answer:
0.218mol
Explanation:
Using Avogadro's law equation;
Va/na = Vb/nb
Where;
Va = volume of gas A
Vb = volume of gas B
na = number of moles of gas A
nb = number of moles of gas B
According to the information in this question,
na = 0.4mol
nb = x mol
Va = 220ml
Vb = 120ml
Using Va/na = Vb/nb
220/0.4 = 120/x
Cross multiply
0.4 × 120 = 220 × x
48 = 220x
x = 48/220
x = 0.218mol
12 grams of carbon is burnt with a certain amount of air containing 36 grams of oxygen. The product contains 24 grams of Co, and 4 grams of CO. Calculate the percentage of excess oxygen.
Answer:
C
Oxygen gas is limiting.
C(s) + O
2
→CO
2
(g)
No. of moles of carbon =
12
36
=3 moles
No. of moles of oxygen =
32
32
=1 moles
So, 2 moles of carbon is left and oxygen will be completed.
So, O
2
is limiting reagent.
Answer:
14.5
Explanation:
not sure how I got it but I hope this helped!
What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.690 M in sodium chloride and 0.08 M in zinc sulfate? Assume complete dissociation for electrolytes.
Answer:
1. Water purification method by reverse osmosis – membrane filtration
2. Method of purifying pure water by filter
3. Demineralization by ion exchange method
3. Demineralization by ion exchange method
Explanation:
· Acids are not safe to be used, but our stomach secretes hydrochloric acid. What would happen if the stomach does not carry out this task? Mark them brainlist
A cyclopropane-oxygen mixture is used as an anesthetic. If the partial pressure of cyclopropane in the mixture is 330 mmHg and the partial pressure of the oxygen is 1.0 atm, what is the total pressure of the mixture in torr
Answer:
1090 Torr
Explanation:
Step 1: Given data
Partial pressure of cyclopropane (pC₃H₆): 330 mmHg (330 Torr)Partial pressure of oxygen (pO₂): 1.0 atmStep 2: Convert pO₂ to Torr
we will use the conversion factor 1 atm = 760 Torr.
1.0 atm × 760 Torr/1 atm = 760 Torr
Step 3: Calculate the total pressure of the mixture (P)
The total pressure of the mixture is the sum of the partial pressures of the gases.
P = 330 Torr + 760 Torr = 1090 Torr
Phosphagens are a group of molecules that includes creatine phosphate (in vertebrates), and arginine phosphate, lombricine, and phosphoopheline (in invertebrates). These molecules have similar functions in different organisms.
a. True
b. False
Answer: True
Explanation:
Phosphagens are high energy storage compounds that are usually found in the tissue of animals.
Based on the question, the molecules have similar functions in different organisms such as the fact that they can accept phosphoryl groups from ATP in a situation where the ATP is in excess. Also, they donate phosphoryl groups to ADP in order for the regeneration of ATP.
what is food nutrients
Answer:
Nutrients arw compounds in foods essential to life and heath
Answer: In simple terms nutrients are the energy that you get from food certain foods give more nutrients and others give close to none. That is what nutrients in your food is
Explanation:
what type of bonding does Sodium Sulphate comes under?and explain in detail please
Answer:
The bond between sodium sulfate is an ionic bond since it's a bond between a metal and non metals however the bond between sulfur and oxygen is a covalent bond since the two are non metals and the other reason that makes this an ionic bond is that there is both losing and gaining of electrons..
I hope this helps
Read the scales of this balance.
The unknown sample has a mass of:
11.2 g
01.012 kg
1.220 g
O 1.200 g
Answer:
and I'll call you when the party's over
quiet when I'm come in home
when I'm all alone
Answer:
Explanation:
Don't you know too much already?
I'll only hurt you if you let me
Call me friend but keep me closer (call me back)
And I'll call you when the party's over
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?
Answer:
1087.84 J
Explanation:
From the question given above, the following data were obtained:
Mass of metal (Mₘ) = 70 g
Temperature of metal (Tₘ) = 80 °C
Mass of water (Mᵥᵥ) = 100 g
Temperature of water (Tᵥᵥ) = 22 °C
Equilibrium temperature (Tₑ) = 24.6 °C
Heat lost by metal (Qₘ) =?
NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Thus, we shall determine the heat gained by water. This can be obtained as follow:
Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
Qᵥᵥ = 100 × 4.184 (24.6 – 22)
Qᵥᵥ = 418.4 × 2.6
Qᵥᵥ = 1087.84 J
Thus, the heat gained by water is 1087.84 J.
Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)
Qₘ = Qᵥᵥ
Qᵥᵥ = 1087.84 J
Qₘ = 1087.84 J
Therefore, the heat lost by the metal is 1087.84 J
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
What is a calorimeter?A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.
Let's use the following expression to calculate the heat absorbed by the water.
Qw = c × m × ΔT
Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ
where,
Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.
Qw + Qm = 0
Qm = -Qw = -10.8 kJ
A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.
Learn more about calorimeters here: https://brainly.com/question/12431493
Activation energy is:
A. The energy needed to begin breaking the bonds of reactants.
B. None of these.
C. The maximum amount of energy reactants can hold.
D. The energy needed to begin breaking the bonds of products.
Activation energy is the energy needed to begin breaking the bonds of reactants. Hence, option A is correct.
What is activation energy?Activation energy is defined as the minimum amount of energy necessary to initiate a chemical reaction.
Hence, activation energy is the energy needed to begin breaking the bonds of reactants.
Learn more about activation energy here:
https://brainly.com/question/2410158
#SPJ5
what is bond? write it's type
Based on the standard EMF series and your knowledge of half-reactions, determine the cell potential and spontanei ty of a cell that consists of a pure cobalt electrode in a solution of Co^2+ ions; the other half is a lead electrode immersed in a Pb^2+ solution.
Pb +2e- Pb Sn +2e Sn Ni 2e Ni Co 2e -0.126 -0.136 -0.250 -0.277 Co
a. +0.403, spontaneous
b. -0.403, nonspontaneous
c. +0.151, spontaneous
d. -0.151, nonspontaneous
Answer:
+0.151, spontaneous
Explanation:
Given that;
Co^2+(aq) + 2e ---->Co(s) -0.28 V
Pb^2+(aq) + 2e ---->Pb(s). -0.13 V
Hence Co is the anode and Pb is the cathode
E°cell = E°cathode - E°anode
So;
E°cell = -0.13 V - (-0.28 V)
E°cell = 0.15 V
The cell reaction is spontaneous since E°cell is positive.
Choose the correct answer:
1.9 × 103 g
1.9 x 106 g
1.9 x 1010 g
Answer:
A. 1.9 × 103 g
(next one)
Which metric unit would be the best choice to report the result?
A. kg
Answer:
1. 2
2. 1.9 × 10^3 g
3. kg
Explanation:
What conversion factor would we need to convert moles of helium to atoms of helium?
Explanation:
The conversion factor required to convert moles of helium to atoms of helium is equal to 6.023*10²³ (NA). Helium is a monoatomic gas. Each molecule of helium gas contains one atom of helium. Therefore, one mole of helium gas will contain one mole atoms of helium.
g consider the following pair of aqueous solutions. which pair will result in the formation of a precipitate? give the formula for the precipitate in the blank. write none if no precipitate forms. a) libr and nh4no3 b) kcl and pb(ch3coo)2
Answer:
kcl and pb(ch3coo)2
The precipitate is PbCl2
Explanation:
Let us take the options provided one after the other;
In the first case, we have;
LiBr(aq) + NH4NO3(aq) ----> LiNO3(aq) + NH4Br(aq)
You can see that no precipitate is formed here.
In the second case;
2KCl(aq) + Pb(CH3COO)2(aq) ----> PbCl2(s) + 2CH3COOK(aq)
The precipitate here is PbCl2.
A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.
Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
[tex]M=58g[/tex]
Explanation:
From the question we are told that:
Heat Capacity [tex]H=0.897[/tex]
Mass of water [tex]M=200g[/tex]
Initial Temperature of Aluminium [tex]T_a=85.6[/tex]
Initial Temperature of Water [tex]T_{w1}=16.0[/tex]
Final Temperature of Water [tex]T_{w2}=16.0[/tex]
Generally
Heat loss=Heat Gain
Therefore
[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]
[tex]M=58g[/tex]
Electrical
City
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Specles Present
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nenelectrolyte
Observation Ne, few, or many
dissociade
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Sugar, CHO..(s)
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اسلو داد
0.IM HNO
0,1M NaOH
0.1M NH4OH
0,1M CHO
0.1M AI(NO3)3
Methanol, CH3OH (aq)
0.1M CuSO4
Answer:
Explanations you mom bumb it’s a:
It’s 2
