Answer:
atom
Explanation:
The sodium atom has a single valence electron that it can easily lose. (If the sodium atom loses its valence electron, it achieves the stable electron configuration of neon.) The chlorine atom has seven valence electrons and can easily gain one electron.
What does IUPAC stand for?
Answer: International Union of Pure and Applied Chemistry
The International Union of Pure and Applied Chemistry (IUPAC), established in 1919, is the international body that represents chemistry and related sciences and technologies.
Commercially prepared cloning vectors such as pUC18 are designed to contain several useful features. An example of one of these features is ________.
Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.
The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.
A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.
These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.
Learn more in:
https://brainly.com/question/14020637
Doing Labs at home
I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!
Answer:
go get the stuff.
Explanation:
Which is larger? 12 milligrams or 12 kilograms
Answer:
12 kilograms is larger.
Explanation:
Of the three units, the kilogram is the largest and the milligram is the smallest. The prefix “kilo” means a thousand and “milli” means one-thousandths. A gram is the basic unit of mass.
What would be the freezing point of a solution that has a molality of 1.506 m which was prepared by dissolving biphenyl (C12H10) into naphthalene
The freezing point of a solution prepared by dissolving biphenyl (C12H10) into naphthalene is 67.99oC.
Freezing point can be obtained from;
ΔT = K m i
ΔT = freezing point depression
m = molality of the solution = 1.506 m
i = Van't Hoff factor = 1 for molecular substances
K = Freezing constant of naphthalene = 8.15 oC/m
ΔT = 8.15 oC/m × 1.506 m × 1 = 12.27oC
Freezing point of pure naphthalene = 80.26 °C
ΔT = Freezing point of pure naphthalene - Freezing point of solution
Freezing point of solution = Freezing point of pure naphthalene - ΔT
Freezing point of solution = 80.26 °C - 12.27oC = 67.99oC
Learn more:
What is the kinetic energy of a mole of Ar atoms moving with a speed of 650m/s
Molar mass of Argon, Ar = 39.95 g Mass of 1 atom, m = 39.95 / (6.023 * 10^23) = 6.64 * 10^(-23) g =6.64 * 10^(-26) Kg Now, Kinetic Energy, KE = (1/2) * m * v^2 where velocity,v = 650 m/s Mass of 1 atom, m = 6.64 *...
- BRAINLIEST answerer
Answer:
1.40×10-20
Explanation:
Oh calculate the mass of 1 atom of argon the molar mass of argon 39.95g that is 6.0²×10²³atoms maybe...?
Your skin is an important organ and has several functions. All of the functions below are performed by the skin EXCEPT
A)
makes vitamin D.
B
prevents-dehydration:
C)
maintains body temperature.
D)
works with bones to help you move
Answer:
D. works with bones to help you move
Explanation:
Your muscles and tendons are what help bones move.
hydrodistillation explain ????
Answer:
Explanation:
Hydrodistillation is a traditional method for the extraction of bioactive compounds from plants. In this method, plant materials are packed in a still compartment then water is added in sufficient amount and brought to a boil. ... The vapor mixture of water and oil is condensed by indirect cooling with water.
radium-223 decays with a half-life of 11.4 days, how long will it take for a 0.240-mol sample of radiuim to decay to 1.88 x 10-3 mol
The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
We'll begin by calculating the number of half-lives that has elapsed.
Original amount (N₀) = 0.240 mole
Amount remaining (N) = 1.88×10¯³ mole
Number of half-lives (n) =?N = 1/2ⁿ × N₀
1.88×10¯³ = 1/2ⁿ × 0.240
Cross multiply
1.88×10¯³ × 2ⁿ = 0.240
Divide both side by 1.88×10¯³
2ⁿ = 0.240 / 1.88×10¯³
2ⁿ = 128
2ⁿ = 2⁷
n = 7Thus, 7 half-lives has elapsed
Finally, we shall determine the time.Number of half-lives (n) = 7
Half-life (t½) = 11.4 days
Time (t) =?t = n × t½
t = 7 × 11.4
t = 79.8 daysTherefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
Learn more: https://brainly.com/question/13266270
hormones that are essential for normal body growth and maturation include all the following except?
a. thyroid hormone
b. growth hormone
c. Ghrelin
d. insulin
What will happen to the temperature of an object if the kinetic energy of the particles increases?
The temperature of an object will increase if the kinetic energy of the particle increases.
Kinetic energy is an energy that is said to be in motion. According to the kinetic molecular theory of ideal gas, the particles of the gas are usually moving in constant random motion and they exert no force on each other.
Also, the temperature varies directly proportional to the average kinetic energy of the gas particles. As a result, when the kinetic energy of the particles increases, the temperature will also increase.
Learn more about the kinetic theory of gas here:
https://brainly.com/question/9949658?referrer=searchResults
What is the formula for sodium phosphate
Answer:
Explanation:
It''s chemical formula is Na3PO4
if 2.4l of chlorine at 400 mm hg are compressed to 725 mm hg at a constant temperature. what is the new volume?
Answer: 1.324L
Explanation: use Boyles law, sorry so late!
Br2(l) + 2Nal(aq) — 12(s) + 2NaBr(aq)
Which elements are oxidized and reduced in the reaction?
(1 point)
O Sodium (Na) is oxidized, and bromine (Br) is reduced.
O Bromine (Br) is oxidized, and iodine (1) is reduced.
O Bromine (Br) is oxidized, and sodium (Na) is reduced.
Olodine (I) is oxidized, and bromine (Br) is reduced.
Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.
To determine which elements are oxidized and reduced,
First, we will define the terms Oxidation and Reduction
Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.
Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.
The given chemical equation is
Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)
Oxidation number of Bromine decreased from 0 to -1.
Therefore, Bromine is reduced.
Oxidation number of Iodine increased from -1 to 0.
Therefore, Iodine is oxidized.
Oxidation number of sodium did not change.
Therefore, Sodium is neither oxidized nor reduced.
Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.
Learn more here: https://brainly.com/question/12913997
In experiment 9, in one operation, we heat up the alcohol with acid and do a concurrent distillation. What was the purpose of doing this
Answer:
we heat up because the component with lower boiling evaporates first,
leaving the other behind
hydrogen bond is a strong dipole interaction for example water and other molecules true or false
Answer:
hydrogen bond is a type of dipole-dipole interaction; it is not a true chemical bond it is a mere electrostatic attraction. These attractions can occur between molecules or within different parts of a single molecule. Intramolecular hydrogen bonds are those which occur within one single molecule.
Explanation:
CAN I GET BRAINLIEST
Select the container from the figure (Figure 1) that represents the dilution of a 4 % (m/v) KCl solution to each of the following: Figure1 of 1 There is a diagram showing several containers. One container is filled with 4 percent of mass to volume solution of NaCl. Container 1 is filled with a solution in which volume is two times less than the volume of NaCl solution. Container 2 is filled with a solution of a volume two times larger compared to the NaCl solution. Container 3 is filled with a solution of a volume two times larger than the volume of the solution in container 2. Part A 2 % (m/v) KCl
From the dilution formula, we have that at constant value of the solute, the volume of the solution is inversely proportional to the concentration.
The correct responses are;
Part A: Container 2Part B: Container 3Reasons:
The given parameters are;
The concentration of the KCl solution = 4% m/v
Taking the solution as solution of KCl
The volume of the solution in container 1 = Two times less than the volume of KCl solution.
[tex]V_{container \, 1} = \displaystyle \mathbf{ \frac{1}{2} \cdot V_{4\% \, solution}}[/tex]
The volume of the solution in container 2 = Two times larger compared to the volume of KCl solution.
[tex]V_{container \, 2} = \mathbf{\displaystyle 2 \times V_{4\% \, solution}}[/tex]
The volume of the solution in container 3 = Two times larger than the container two solution volume.
[tex]V_{container \, 3} = \displaystyle \mathbf{ 2 \times V_{container \, 2}}[/tex]
Therefore;
[tex]V_{container \, 3} = \displaystyle 2 \times 2 \times V_{4\% \, solution} = \mathbf{4 \times V_{4\% \, solution }}[/tex]
Part A Required:
a. To select the container that represent the dilution of the 4% solution to 2%
Solution:
The dilution formula is; C₁·V₁ = C₂·V₂
Therefore;
[tex]\displaystyle V_1 = \mathbf{\frac{C_1 \cdot V_1}{C_2}}[/tex]
C₁ =4%, C₂ = 2%, we get;
[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{2} = 2 \cdot V_1[/tex]
The volume of the container that represents a 2% dilution is container 2
which is filled with a solution of a volume two times larger compared to the
KCl solution.
Part B:
Required:
The container diluted to a 1% m/v KCl solution.
Solution;
Using the dilution formula, we have;
C₁ = 4%, C₂ = 1%
Therefore;
[tex]\displaystyle V_1 = \frac{C_1 \cdot V_1}{C_2}[/tex]
[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{1} = \mathbf{4 \cdot V_1}[/tex]
The volume of the solution is four times the volume of the 4% KCl solution, which is equivalent to the volume in container 3.
Possible parts of the question are;
Select the container that represents the dilution of the 4% (m/v) KCl solution to obtain the solutions that follows;
Part A: a 2% (m/v) KCl solution
Part B: a solution that is a 1% (m/v) KCl solution
Please see attached drawings
Learn more here:
https://brainly.com/question/11493179
The pH of an acidic solution is 4.83. What is [H"]?
[tex]pH = -\log[H^{+}] \\\\\implies \log[H^{+}] = -pH\\\\\implies [H^{+}] = 10^{-pH}\\\\\implies [H^{+}] = 10^{-4.83} = 0.000015[/tex]
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer:
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer: Tristan is most like the Golgi Body
Explanation:
which mass movement does this describe?
Answer:
landslide mudflows slump and creep
Explanation:
no explanation
What question would a student need to ask to form a compound with Group 16 nonmetals
A.
Will group 16 elements lose electrons to bond with group 2 in an XY format?
B.
Will group 16 elements gain electrons to bond with group 1 in an XY2 format?
C.
Will group 16 elements gain electrons to bond with group 2 in an XY format?
D.
Will group 16 elements lose electrons to bond with group 1 in an X2Y format?
Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.
If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Learn more: https://brainly.com/question/14281129
A 0.48-mole sample of helium gas occupies a volume of 11.7 L . What is the volume of 0.72 mol of helium gas under the same conditions
AnswerExplanation:I finsished d
Answer:
17.55L
Explanation:
0.48mol : 11.7
0.72mol : x
0.48x = 8.424
x = 17.55
which of the following are compounds (select all that apply)
a) Br2
b)NO2
c) KBr
d) Fe
Answer:
KNO2, KBr
Explanation:
Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.
Using the combined gas
law, what is the final
volume of the Helium gas?
This is the answer i hope it helps
A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?
HELP ME OUT PLEASE!!!!
Which statement correctly describes one of the changes?
A) Picture I shows a chemical change, because a new substance is formed.
B) Picture Il shows a chemical change, because a new substance is formed.
C) Picture I shows a chemical change, because the same substance changes form,
D) Picture Il shows a chemical change, because the same substance changes form.
Answer:
Answer D. Picture II shows a chemical change, because the same substance changes form
Explanation:
This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal
Answer:
D Im 90% Sure
Explanation:
If its not right i owe you one I did this one before
Which is true of protons and neutrons?
1. They have approximately the same mass and the same charge.
2) They have approximately the same mass but different charge.
by The have different mass and different charge.
O sette
4) They have different mass but the same charge.
Answer:
[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
Name the following Type 1 compounds:
1. Naci
2. KI
3. Cas
4. CsBr
5. Mgo
6. CSF
7. AICI:
8. Mg2
9. Rb20
10.Sr12
11.KAS
Answer:
44
Explanation:
Magnesium metal burns in air with an intense bright light according to the equation
2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ
A.) What is the amount of energy in kJ produced when 4.5 mol of Mg is burned in the presence of excess oxygen?
The heat produced by 4.5 moles of magnesium when burnt is 2700 kJ.
A thermochemical reaction is a reaction in which the amount of heat lost or gained is included in the reaction equation. The thermochemical reaction equation for the combustion of magnesium is shown as follows;
2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ
From the reaction equation;
2 moles of magnesium produced 1200 kJ of heat
4.5 moles of magnesium will produce 4.5 moles × 1200 kJ/2 moles
= 2700 kJ
Learn more: https://brainly.com/question/14281129
Who wants to simp for me??
Answer:
qrtyuioplkjhgfdssssssazxcvbn
Which option correctly describes the relative charges and masses of the subatomic particles?
Answer:
D
Explanation:
D Is The Answer Fella, My Head Hurt Really Bad