Answer:
True
Explanation:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
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.
what is bond? write it's type
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
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.
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
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
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
Which redox reaction would most likely occur if zinc and copper metal were
added to a solution that contained zinc and copper ions?
Click for a reduction potential chart
A. Cu + Zn → Cu2+ + Zn2+
B. Cu + Zn2+
Cu2+ + Zn
C. Cu2+ + Zn → Cu + Zn2+
D. Cu2+ + Zn2+ → Cu + Zn
Answer:
C
Explanation:
b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex] is the redox reaction. Hence, option C is correct.
What is Redox Reaction?A chemical reaction taking place between an oxidizing substance and a reducing substance.
The oxidizing substance is used to lose electrons in the reaction, and the reducing substance is used to gain electrons.
On the reduction potential chart, zinc is a stronger oxidizing agent than, Copper (Cu), which is a reducing agent as compared to silver
The redox reaction most likely occurs if silver and copper metal were added to a solution that contained silver and copper ions is ;
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex]
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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.
the ability for carbon to form long chain or rings is
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!
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0,1M CHO
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Answer:
Explanations you mom bumb it’s a:
It’s 2
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
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 process occurs during the corrosion of iron?
Answers
A.
Iron is oxidized.
B.
Iron is reduced.
C.
Iron (III) is oxidized.
D.
Iron (III) is reduced.
Answer:
A
Explanation:
The iron corrodes so it oxidized
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.
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Which substance would be the most soluble in gasoline?
Select one:
A. hexane
B. NaNO3
C. HCI
D. water
E. Nacl
I think the answer most be d
In chemistry like dissolves like hence hexane will dissolve in gasoline.
Dissolution stems from intermolecular interaction between solute and solvent molecules.
If this interaction can not occur, dissolution of one substance in another is impossible.
Hexane dissolves in gasoline because the both substances are non-polar and can interact with each other effectively.
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Arrange the following compounds in order of increasing reactivity (least reactive first.) to electrophilic aromatic substitution:.
Bromobenzene Nitrobenzene Benzene Phenol
a. Bromobenzene < Nitrobenzene < Benzene < Phenol
b. Nitrobenzene < Bromobenzene < Benzene < Phenol
c. Phenol < Benzene < Bromobenzene < Nitrobenzene
d. Nitrobenzene < Benzene < Bromobenzene < Phenol
Answer:
Nitrobenzene < Bromobenzene < Benzene < Phenol
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.
Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.
However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.
Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;
Nitrobenzene < Bromobenzene < Benzene < Phenol
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:
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.
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Based on the following observations decide the order of reactivity for hydrogen, magnesium, and copper. Hydrochloric acid reacts with magnesium but did not react with copper. magnesium reacted with copper sulfate. Write your answers in the blanks. For magnesium write magnesium for hydrochloric acid write hydrogen and for copper sulfate write copper.
Answer:
Mg> H> Cu
Explanation:
We can see from the question that hydrochloric acid reacted with magnesium as follows;
Mg(s) + 2HCl(aq) ----> MgCl2(aq) + H2(g)
Copper does not react with HCl which means that copper is less reactive than hydrogen hence it can not displace hydrogen from a dilute acid solution.
The order of reactivity of the elements then is ; Mg> H> Cu
What trends were seen in medeleevs periodic table
Answer:
groups are based on how many electrons to become stable
Explanation:
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]
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.
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
A 10.00 mL sample of ethanol is added to 250.0 mL of water in order to observe the freezing point depression of the water. Consider the possible errors in this experiment, and determine whether each would cause the freezing point of the water to increase, decrease, or remain the same compared with the expected results. The ethanol sample is left out for several minutes before being added to the water, allowing a Choose... significant amount to evaporate. Some of the water is splashed out of the container before the ethanol is added. Choose... Choose... Instead of starting the experiment with room temperature water, the experiment is started with cold water. It is discovered that the 10.00 g ethanol sample is actually 95% ethanol and 5% water. Choose... Choose... Some of the ethanol and water solution is spilled out of the container during temperature measurement
Answer:
b
Explanation:
Addition of a solute lowers the freezing point of a substance.
Let us recall that the addition of a solute to a solution lowers the freezing point of the pure solution. This is because, freezing point depression is a colligative property. It depends on the number of solute molecules present.
The following are possible errors in the experiment:
The ethanol sample is left out for several minutes before being added to the water allowing a significant amount to evaporate. This causes the freezing point of water to increase compared with the expected results.Some of the water is splashed out of the container before the ethanol is added. This causes the freezing point of water to increase compared with the expected results. It is discovered that the 10.00 g ethanol sample is actually 95% ethanol and 5% water. This causes the freezing point of water to decrease compared with the expected results.Some of the ethanol and water solution is spilled out of the container during temperature measurement. This causes the freezing point of water to remain the same compared with the expected results.Learn more about colligative properties:https://brainly.com/question/8073802
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:
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.
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
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.
