Stalactites and stalagmites form as ________ precipitates out of the water evaporating in underground caves.
Group of answer choices
hydrochloric acid
sodium bicarbonate
calcium carbonate
sodium chloride
sodium hydroxide
Answer:
calcium carbonate
Explanation:
A stalactite is an icicle-looking mould that is formed by the precipitation of natural minerals as a result of water dripping from the ceiling, hanging from a cave.
A stalagmites in the other hand, grows upwards and is also a mound that is formed by the deposits of minerals gotten by the water dripping on the floor of a cave.
Therefore, stalactites and stalagmites form as calcium carbonate precipitates out of the water evaporating in underground caves.
Question 9 of 25
How many hydrogen atoms are in a molecule of table sugar (C12H,2011)?
O A. 12
B. 45
C. 11
D. 22
SUBMIT
D.22
is my answer than welcome
what is Lewis acid and Lewis base? give examples
Explanation:
example is copper iron...........
A gas occupies a volume of 202 ml at a pressure of 505 torr. To what pressure must the gas be subject in order to change the volume to 65.0 ml
Answer:
1569 torr
Explanation:
Assuming ideal behaviour and constant temperature, we can solve this problem by using Boyle's law, which states:
V₁P₁ = V₂P₂Where in this case:
V₁ = 202 mLP₁ = 505 torrV₂ = 65.0 mLP₂ = ?We input the data given by the problem:
202 mL * 505 torr = 65.0 mL * P₂And solve for P₂:
P₂ = 1569 torrAssuming tea leaves contain 5.0% caffeine by weight what is the maximum weight of caffeine you could isolate from 10.g of tea leaves? Show your work.
Answer:
0.50 g Caffeine
Explanation:
Step 1: Given data
Concentration of caffeine by weight in tea leaves: 5.0%
Mass of tea leaves: 10. g
Step 2: Calculate the maximum weight of caffeine that can be isolated
The concentration of caffeine by weight in tea leaves is 5.0%, that is, there are 5.0 g of caffeine per 100 g of tea leaves. The maximum weight of caffeine in 10. g of tea leaves is:
10. g Tea leaves × 5.0 g Caffeine/100 g Tea leaves = 0.50 g Caffeine
You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Answer:
The pH is greater than 7 at the equivalence point.
Explanation:
Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.
When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.
Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.
A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.
What are weak acids?Weak acids are acids which only ionize partially in aqueous solutions.
When weak acids are dissolved in water, they produce only few hydrogen ions.
A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.
The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.
Learn more about equivalence point at: https://brainly.com/question/18933025
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
3) Convert 0.250 moles of aluminum sulfate to grams.
4) Convert 2.70 grams of ammonia to moles.
Answer:
0.000731 grams aluminium sulfate
46.0 mols ammonia
Explanation:
ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol
[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]
NH3 has a molar mass of 17.031 g/mol
[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]
Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]
we have to find the 0.250 moles of aluminum sulphate.
[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]
[tex]\\\\\\[/tex]
Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]
We have to find 2.70 grams of ammonia
[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]
examples s name of thosse food items we can store for a month?
Answer:
1. Nuts
2. Canned meats and seafood
3. Dried grains
4. Dark chocolate
5. Protein powders
When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net ionic equation for the reaction.
Answer:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:
[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]
Now, we can separate the nitrates in ions as they are aqueous to obtain:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]
And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Best regards!
CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.
Answer:
[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
[tex]m=552.4g-464.7g=87.7g[/tex]
So that we are now able to calculate the density in g/mL first:
[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]
Now, we proceed to the conversion to lb/in³ by using the following setup:
[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Regards!
How many atom in protons
Answer:
Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).
repining of fruits is which type of change
Answer:
irreversible.
I hope this will help you
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
办理教留服学位学历认证Q/微29304199英属哥伦比亚UBC毕业证文凭学位证书offer操办英属哥伦比亚留信认证成绩单
Answer:
please translate in english
How many colors are there in a rainbow?
[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]
There are 7 colours in a rainbow The colours of the rainbow are Red, Orange, Yellow, Green, Blue, Indigo and Violet.Explanation:
there r seven colors in a rainbow.red, orange, yellow, green, blue, indigo and violet.hope it helps.stay safe healthy and happy..why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
Explain the term global warming
Answer: A gradual increase in the overall temperature of the earth's atmosphere generally attributed to the greenhouse effect caused by increased levels of carbon dioxide, chlorofluorocarbons, and other pollutants.
Explanation:
Choose the correct answer to make the statement true.
a. An exothermic reaction has a positive ΔH and absorbs heat from the surroundings.
b. An exothermic reaction feels warm to the touch. a positive ΔH and gives off heat to the surroundings.
c. An exothermic reaction feels warm to the touch. a negative ΔH and absorbs heat from the surroundings.
d. An exothermic reaction feels warm to the touch. a negative ΔH and gives off heat to the surroundings.
e. An exothermic reaction feels warm to the touch.
Is sucrose classified as aldose or ketose?
Answer:
Because sucrose is a complex disaccharide, it is not classified as either an aldose or a ketone. Instead, it is a compound that contains both. glucose is aldose sugar and fructose is a ketose sugar.
what is the hybridisation of the central carbon in CH3C triple bonded to N
Explanation:
the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet
What is the direct function of the energy released from the nuclear chain reaction in a nuclear power plant? turning the blades of the turbine heating water to produce steam powering the condenser carrying electricity from the plant to consumers
Answer:
the energy released is to make steam to create electricity. yes you are right i just didnt feel like being super technical
In a nuclear reaction, the direct function of the energy released from the nuclear chain reaction in a nuclear power plant is turning the blades of the turbine heating water to produce steam .
What are nuclear reactions?There are two types of nuclear reactions which are nuclear fusion and nuclear fission .They involve the combination and disintegration of the element's nucleus respectively.
In nuclear fission, the nucleus of the atom is bombarded with electrons of low energy which splits the nucleus in to two parts .Large amount of energy is released in the process.It is used in nuclear power reactors as it produces large amount of energy.
In nuclear fusion,on the other hand, is a reaction which occurs when two or more atoms combine to form a heavy nucleus.Large amount of energy is released in the process which is greater than that of the energy which is released in nuclear fission process.
Learn more about nuclear reactions,here:
https://brainly.com/question/12786977
#SPJ7
sự sắp xếp nguyên tử trong vật chất
Answer:
sosksjsjjs
Explanation:
even i know how to type şïllily
The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chloride dissolves in water. Be sure to specify states such as (aq) or (s).
Answer:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Explanation:
Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Help naming this plzzzzzzzzzzzzz
Answer:
A. 3-chloro-1-methylcyclobutane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).
Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.
Regards!
2. For each of the ionic compounds in the table below, name the compound and explain the rule that you
used in formulating your name for the compound.
Name:
Rule for naming compound:
-PbF4
-NH4NO3
-Li2S
Answer:
2
Explanation:
Lead(|V) fluoride
Ammonium Nitrate
Lithium sulfide
For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.
The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.
The second one is just two polyatomic ions which you just have to remember.
The last one is the typical ionic compound naming technique i guess.
A ligand is a molecule or ion that acts as a
Answer:
Lewis base/electron pair donors
Explanation:
Ligands are ions or neutral molecules which bond together with a central ion. They act as election pair donors, also known as Lewis bases, while the central ion they are connected to acts as the acceptor.
Therefore, a ligand is a molecule or ion that acts as a Lewis base/electron pair donors
A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment, she recovers 2.775 g of sand and 0.852 g of salt.a. What was the percent composition of sand in the mixture according to the student's data? b. What was the percent recovery?
Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
Plastic is a polymer
-True
-False
TRUE
Explanation:
*not sure about this answer
Calculate the concentration of ammonium nitrate in a solution prepared by dissolving 3.20 g of the salt in enough water to make 100. mL of solution, then diluting 2.00 mL of this solution to a volume of 25.00 mL.
Answer:
.032 M .
Explanation:
Molecular weight of ammonium nitrate is 80 .
3.2 g = 3.2 / 80 moles
= .04 moles
volume = 100 mL = 0.1 L
Molarity of 100 mL solution = .04 moles / 0.1 L
= 0.4 M solution.
Now 2 mL solution of 0.4 M is diluted to a volume of 25 mL .
Using the formula S₁ V₁ = S₂V₂
0.4 M x 2 mL = S₂ x 25 mL
S₂ = .4 x 2 / 25
= .032 M
Hence required concentration is .032 M .