Answer:
Vanlency of hydrogen - 11
Electrons of hydrogen - 1
Answer:
The answer is: 1
Hope this helps :) <3
Explanation:
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
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
the ability for carbon to form long chain or rings is
How many moles of HNO3 are needed to prepare 5.7 liters of a 1.64 M solution of HNO3
We are given:
Volume of Solution: 5.7 Liters
Concentration: 1.64 Molar
Number of moles required:
We know that the formula for concentration (in molar) is:
M = N / V
where M is the concentration, N is the number of moles, and V is the volume of solution (in L)
plugging the given values in the equation:
1.64 = N / 5.7
N = 1.64 * 5.7
N = 9.35 moles
(d) 40g of sulphur
Calculate the number of moles of 40g of sulphur
Answer:
It is 1.25 moles
Explanation:
Molar mass of sulphur = 32 g
[tex]{ \bf{moles = \frac{given \: mass}{molar \: mass} }} [/tex]
Substitute:
[tex]{ \sf{moles = \frac{40}{32} }} \\ { \sf{ = 1.25 \: moles}}[/tex]
A substance is tested and has a pH of 7.0. How would you classify it?
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
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?
Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]
[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]
[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]
Hence, the percent abundance of O-18 is:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]
[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]
Hence, the percent abundance of oxygen-18 is:
[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]
Therefore, the percent abundance of oxygen-18 is 1.9066%.
I hope it helps you!
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
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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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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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
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.
2. Calculate the wavelength of the emitted photon from hydrogen for the transition from ni = 3 to nf = 2. What part of the visible spectrum is this wavelength? Visible wavelengths are: Red 700 - 620 nm, Yellow 620 - 560 nm, Green 560 - 500 nm, Blue 500 - 440 nm, and Violet 440 - 400 nm.
Answer:
The correct answer is "654.54 nm".
Explanation:
According to the question,
⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]
By substituting the values, we get
[tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]
[tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]
[tex]=1.1\times 10^7(\frac{5}{36} )[/tex]
[tex]=654.54\ nm[/tex]
Thus the above is the right solution.
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]
If a hydrogen of an alkane is replaced by NH, the compound becomes_________
a. alcohol
b. carboxylic acid
c. phenol
d. amine
Answer:
d. amine.
It becomes an amine.
Explanation:
With general formular
[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]
R is the aryl group such as alkane
What trends were seen in medeleevs periodic table
Answer:
groups are based on how many electrons to become stable
Explanation:
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
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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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.
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.
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
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:
Draw the structure of the organic product(s) of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide, followed by aqueous workup. You do not have to consider stereochemistry. If a compound is formed more than once, add another sketcher and draw it again. Alternatively, you may use the square brackets tool to add stoichiometries greater than one. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple products using the sign from the drop-down menu.
Answer:
See explanation and image attached
Explanation:
The product of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide is triphenyl methanol.
The reaction proceeds by nucleophillic reaction as the carbonyl moiety is attacked. A tetrahedral intermediate is formed. Loss of the -OMe group is accompanied by the attack of the first molecule of PhMgBr.
Attack by a second PhMgBr molecule yields trimethyl phenoxide. Protonation of this specie yields the final product which is obtained by aqueous workup.
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.
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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The enthalpy of formation for CO2 (s) and CO2 (g) is: -427.4 KJ/mole and -393.5 KJ/mole, respectively. The sublimation of dry ice is described by CO2 (s) → CO2 (g).
The enthalpy needed to sublime 986 grams of CO2 is:
(a) 181.5 Kcal
(b) 611.7 Kcal
(c) 248.3 Kcal
(d) 146.2 Kcal
Answer:
a. 181.5 kcal
Explanation:
Step 1: Calculate the enthalpy of the process (ΔH°).
Let's consider the following process.
CO₂(s) → CO₂(g)
We can calculate the enthalpy of the process using the following expression.
ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)
ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))
ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ
According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.
Step 2: Convert 986 g of CO₂ to moles
The molar mass of CO₂ is 44.01 g/mol.
986 g × 1 mol/44.01 g = 22.4 mol
Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂
22.4 mol × 33.9 kJ/1 mol = 759 kJ
We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.
759 kJ × 1 kcal/4.184 kJ = 181.5 kcal
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
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
A reaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is charged with 6.54 kg of H2 and excess N2. A total of 30.4 kg of NH3 are produced. What is the percent yield of the reaction
Explanation:
The given data is:
The mass of hydrogen is 6.54 kg.
The actual yield is 30.4 kg.
The balanced chemical equation of the reaction is:
[tex]N_2(g)+3H_2(g)<=>2NH_3(g)[/tex]
At first the theoretical yield should be calculated by using the balanced chemical equation:
3 mol. of hydrogen forms ---- 2 mol. of ammonia.
The molar mass of hydrogen is 2.0 g/mol.
The molar mass of ammonia is 17.0 g/mol.
Hence, the above statement can be rewritten as:
6g of hydrogen forms --- 34g of ammonia.
Then,
6.54g of hydrogen forms :
[tex]6.54 kg x 34 g / 6 g\\=37.1 kg[/tex]
% yield = (actual yield /theoretical yield )x 100
=(30.4 kg /37.1 kg )x100
=81.9
Hence, % yield is 81.9.
