Answer:
The best existing methods of control—artemisinin-based drug treatment and insect control with chemical sprays and treated bednets—can reduce the burden of disease substantially, and can even eliminate the disease in some regions,
hope this will help you more
What do phase diagrams demonstrate?
Select all that apply.
how the melting and boiling points of a substance change with pressure
how much heat will increase the temperature of a gram of a substance by 1°C
the pressure and temperature conditions in which a substance exists as a liquid, solid, or gas
a comparison of the temperature of a substance to the heat absorbed over time
Answer:
this answer is for first one
Explanation:
A higher evalations, where the atmospheric pressure is much lower, the boiling point increase with increased pressure up to the critical point where the gas and liquid properties become identical
I want to create water out of 45.4 Liters of Oxygen at STP. How much water will I produce?
STP: Standard Temperature and Pressure
72g H2O
36g H20
9g H20
18g H20
Answer:
72.96 of water produce by 45.4 L of oxygen at STP.
Explanation:
[tex]H_2+\frac{1}{2}O_2\rightarrow H_2O[/tex]
1 mole of oxygen=22.4 L at STP
[tex]\frac{1}{2}[/tex]\mole of oxygen=22.4/2=11.2 L
11.2 L of oxygen required to produce water=1 mole
1 L of oxygen required to produce water=1/11.2 mole
45.4 L of oxygen required to produce water=[tex]\frac{1}{11.2}\times 45.4[/tex]
45.4 L of oxygen required to produce water=[tex]\frac{45.4}{11.2}[/tex]moles
1 mole of water=18 g
[tex]\frac{45.4}{11.2}[/tex]moles of water=[tex]18\times \frac{45.4}{11.2}[/tex]
[tex]\frac{45.4}{11.2}[/tex]moles of water=72.96 g
Hence, 72.96 of water produce by 45.4 L of oxygen at STP.
The speed of sound depends on the __?_____ and ____?____ of the medium through which it travels
Answer:
Density and rigidity
The number of moles of aluminum that react with 15.7 grams of oxygen according to the equation 4 Al + 3 0 2 - 2 Al 20 3
Answer:
Explanation:
1mol of [tex]O_2=2*16{gr\over{mol}}=32{gr\over{mol}}\\\\15.7gr->15.7gr*{1\over{32{gr\over{mol}}}}=0.491mol~of~O_2[/tex]
as 1mol of molecular oxigen reacts with 4 mol of aluminium
1 mol of O2 -----------------------------> 4 mol of Al
0.491 mol of O2 ------------------------------> x
[tex]x={0.491*4\over{1}}~mol~of~Al=1.9625~mol~of~Al[/tex]
Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
Calculate the average atomic mass/given their percent abundance
Answer:
A = 65.46 u
Explanation:
Given that,
The composition of zinc is as follows :
Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
We need to find the average atomic mass of the given element. It can be solved as follows :
[tex]A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u[/tex]
So, the average atomic mass of zinc is 65.46 u.
In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5.00 g of vanillin are added to 1 L of water at 25 C?
Answer:
The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.
Explanation:
The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:
MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol
If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:
0.070 mol x 152 g/mol = 10.64 g
Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.
By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?
Answer:
By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?.
Explanation:
The freezing point of a pure solvent decreases further by adding a nonvolatile solute.
This is called depression in freezing point.
When an ionic solute is dissolved then the depression in the freezing point is proportional to the number of ions present in the solution.
In aluminum sulfate, there are five ions formed as shown below:
[tex]Al_2(SO_4)_3(aq)->2Al^3^+(aq)+3SO_4^2^-(aq)[/tex]
But sucrose is a covalent compound and it does not undergo dissociation.
Hence, aluminum sulfate decreases the freezing point of water by five times compared to sucrose.
Explanation:
Vocabulary: dipole, dipole-dipole force, dipole-induced dipole force, electronegativity, intermolecular force, ionic bond, London dispersion force, molecule, nonpolar, nonpolar covalent bond, partial charges, polar, polar covalent bond, valence electron Prior Knowledge Questions (Do these BEFORE using the Gizmo.) 1. A big bully is having a tug-of-war with a small child. There is a ball attached to the middle of the rope. Toward whom will the ball move
Answer:
Towards the big bully
Explanation:
If a big bully and a small child are involved in a thug of war, it is clear that the bully is stronger than the child and he/she will pull the rope used in the thug of war with a greater force.
By so doing, the ball attached at the centre of the rope will naturally be drawn towards the stronger bully.
A technical machinist is asked to build a cubical steel tank that will hold "265" L of water. Calculate in meters the smallest possible inside length of the tank. Round your answer to the nearest .
Answer:
0.64 m
Explanation:
Given that;
1L = 0.001 cubic metre
Then;
263 L = 263 L × 0.001 cubic metre/1L
= 0.263 cubic metre
Volume of a cube = l^3
l= 3√V
l= 3√0.263 cubic metre
l= 0.64 m
An unknown compound's 2,4-DNP product melting range is between 144-146'C. It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test. What would this compound be
Answer:
An unknown compound's 2,4-DNP product melting range is between 144-146'C. It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test. What would this compound be
Explanation:
From the given data it is clear that the unknown compound gives positive test with 2,4-DNP reagent.
That means it has a carbonyl group.Either aldehyde or ketone.
It does not give a silver mirror on the Tollens test and is slow to react to the chromic acid test.
That means aldehyde is absent.
So, the other carbonyl group that is ketone is present in the given unknown compound.
if a bottle of vinegar has 4.78g of acetic acid (CH3COOH) per 100.0 g of solution (mixed with water, what is the molarity of the vinegar? Density of the solution is 1.00g/mL.
Answer:
0.796 M
Explanation:
Step 1: Given data
Gravimetric concentration (Cg): 4.78 g%g
Density of the solution (ρ): 1.00 g/mL
Step 2: Calculate the volumetric concentration of the solution (Cv)
We will use the following expression.
Cv = Cg × ρ
Cv = 4.78 g%g × 1.00 g/mL = 4.78 g%mL
Step 3: Calculate the molarity of the solution (M)
The volumetric concentration is 4.78 g%mL, that is, there are 4.78 g of acetic acid per 100 mL of solution. We can calculate the molarity using the following expression.
M = mass solute / molar mass solute × liters of solution
M = 4.78 g / 60.05 g/mol × 0.1 L = 0.796 M
0.300 moles of sodium nitrite are needed for a reaction. The solution is 0.450 M. How many mL are needed
Answer:
The solution is 0.450 M. How many mL are needed?
- 0.667 mL
Explanation:
2. What is the conductivity of lead as compared to silver?
O A. 7
OB. 15
O C. 30
O D. 61
1. A positive value for AH for a chemical reaction is an indication of which of the following?
A. Reactants have higher potential energy and the reaction is endothermic.
B. Reactants have higher potential energy, and the reaction is exothermic.
C. Products have higher potential energy, and the reaction is endothermic.
D. Products have higher potential energy, and the reaction is exothermic.
Answer:
C. Products have higher potential energy, and the reaction is endothermic.
For the molecular compound dioxygen difluoride , what would you multiply "grams of O2F2 " by to get the units "molecules of O2F2 "
Answer:
Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules / one mole of O₂F₂)
Explanation:
The Avogadros constant gives the the number of specified entities in one mole of a substance. One mole of any substance contains 6.02 × 10²³ particles. Therefore, one mole of O₂F₂ contains 6.02 × 10²³ molecules.
Also, the molar mass of a substance is the mass in grams of one mole of that substance. It is obtained by summing the relative atomic masses of all the atoms of the elements in the substance. For O₂F₂, the molar mass = (2 × 16 + 2 × 19) g/mol = 70 g/mol
Converting to molecules of O₂F₂;
To convert from grams of a substance to molecules of that substance, multiply by the ratio of one mole and mass of one mole, and then by the number of molecules per mole.
Molecules of A = mass of A × (1 mole / mass of one mole) × (6.02 × 10²³ molecules / 1 mole)
Therefore,Molecules of O₂F₂ = mass of O₂F₂ × (1 mole O₂F₂ / 70 g O₂F₂) × (6.02 × 10²³ molecules /one mole of O₂F₂)
Be sure to answer all parts.
Calculate the molarity of each of the following solutions:
(a) 28.5 g of ethanol (C2H5OH) in 4.50 x 102 mL of solution
M M
es
(b) 21.6 g of sucrose (C12H22011) in 67.0 mL of solution
M
(c) 6.65 g of sodium chloride (NaCl) in 96.2 mL of solution
M
Answer: The molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Explanation:
Molarity is the number of moles of a substance present in liter of a solution.
And, moles is the mass of a substance divided by its molar mass.
(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol[/tex]
Now, molarity of ethanol solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M[/tex]
(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol[/tex]
Now, molarity of sucrose solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M[/tex]
(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol[/tex]
Now, molarity of sodium chloride solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M[/tex]
Thus, we can conclude that the molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
o Calculate the pH of 500 ml of buffer solution containing .2M of ammonium sulphate and 0.3M ammonium hydroxide which 5 2.1% ionized in dilute solution. (kb of ammonium hydroxide is 1.8×10^-5)
Answer is 9
pKb=−logK
b=−log10^-5=5
A mixture of ammonium hydroxide and ammonium chloride forms a basic buffer solution.Henderson's equation for the basic buffer solution is as given below-pOH=pKb+log[acid]*[salt]
Substitute values in the above expression.
pOH=5+log0.1*0.1=5
Hence, the pH of the solution is pH=14−pOH=14−5=9
What do you mean by pH ?potential of hydrogena measure of the acidity or alkalinity of a solution equal to the common logarithm of the reciprocal of the concentration of hydrogen ions in moles per cubic decimetre of solution. Pure water has a pH of 7, acid solutions have a pH of less than 7, and alkaline solution with a pH greater than 7.
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Many home barbeques are fueled with propane gas (C3H8)(C3H8). Part A What mass of carbon dioxide is produced upon the complete combustion of 27.9 LL of propane (the approximate contents of one 5-gallon tank)
Answer:
41264 g of CO₂
Explanation:
Combustion reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.
In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.
We have the volume of propane but we need moles of it, so we need to apply density.
Density = mass / volume so mass = density . volume.
Density of propane is: 493 g/L
Mass of propane is 493 g/L . 27.9L = 13754.7 g
We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles
According to reaction, 1 mol of propane can produce 3 moles of CO₂
Our 312.6 moles will produce 312.6 . 3 = 937.8 moles
We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g
Which statement BEST describes the strength of ionic and covalent bonds
Ionic bonds are the strongest type of chemical bond and, therefore, most compounds remain solid with very high melting points
The statement best describes the strength of ionic and covalent bonds is that ionic bond is the strongest bond and has very high melting point.
What are different types of bonding in chemistry?A chemical bond is a lasting attraction between atoms or ions that enables the formation of molecules, crystals, and other structures. The bond may result from the electrostatic force between oppositely charged ions as in ionic bonds, or through the sharing of electrons as in covalent bonds.
Types of Chemical Bonds includes-
Ionic Bonds.
Covalent Bonds.
Hydrogen Bonds.
Polar Bonds.
Generally, ionic bonds are much stronger than covalent bonds. In ionic bonds, there is complete transfer of electrons between elements to form a stable compound. While in covalent bond, there is only sharing of electrons between two elements to form a stable compound.
Therefore, The statement best describes the strength of ionic and covalent bonds is that ionic bond is the strongest bond and has very high melting point.
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Construction of battery cells
Answer:
These batteries are constructed of several single cells connected in series each cell produces approximately 2.1 volts. ... A battery cell consists of two lead plates a positive plate covered with a paste of lead dioxide and a negative made of sponge lead, with an insulating material (separator) in between.
I mix together 50.0 mL of 0.100 M NaIO3, 50.00 mL of 0.100 M NaOH, and 10.0 mL of 0.100 M HIO3. What is the pH of the mixture
Answer:
pH = 12.66
Explanation:
The HIO3 reacts with NaOH as follows:
HIO3 + NaOH → H2O + NaIO3
The moles of HIO3 and NaOH added are:
Moles HIO3:
0.0100L * (0.100mol / L) = 0.00100 moles HIO3
Moles NaOH:
0.05000L * (0.100mol / L) = 0.00500 moles NaOH
As moles NaOH > Moles HIO3, the moles of NaOH that remain are:
0.00500mol - 0.00100mol = 0.00400 moles NaOH.
After the reaction you will have only NaOH and NaIO3. As NaIO3 is a salt, the pH of the solution is determined by only NaOH. Its concentration is:
Moles NaOH: 0.00500 moles NaOH
Volume: 50.0mL + 50.0mL + 10.0mL = 110.0mL = 0.110L
Molarity: 0.0455M NaOH = [OH-]
pOH = -log [OH-] = 1.34
pH = 14 - pOH
pH = 12.66pH is the measure of the hydrogen or the hydronium ion in an aqueous solution. The pH of the mixture containing sodium hydroxide is 12.66.
What is pH?pH is the potential of the hydrogen and is given as a negative log of the hydrogen concentration in the aqueous solution.
The balanced chemical reaction can be shown as:
[tex]\rm HIO_{3} + NaOH \rightarrow H_{2}O + NaIO_{3}[/tex]
Moles of iodic acid are calculated as:
[tex]\begin{aligned} \rm moles &= \rm molarity \times volume\\\\&= 0.100 \;\rm M \times 0.0100 \;\rm L \\\\&= 0.00100\;\rm moles\end{aligned}[/tex]
Moles of sodium hydroxide are calculated as:
[tex]\begin{aligned} \rm moles &= \rm molarity \times volume\\\\&= 0.100 \;\rm M \times 0.05000 \;\rm L \\\\&= 0.00500\;\rm moles\end{aligned}[/tex]
The remaining moles of sodium hydroxide are 0.00500mol - 0.00100mol = 0.00400 moles.
The pH of the mixture will be determined by sodium hydroxide, as sodium iodate is a salt.
The molar concentration of sodium hydroxide is calculated as:
[tex]\begin{aligned} \rm M &= \rm \dfrac{moles}{volume}\\\\&= \dfrac{0.00500}{0.110}\\\\&= 0.0455 \;\rm M\end{aligned}[/tex]
pH is calculated as:
[tex]\begin{aligned} \rm pOH &= \rm -log[OH^{-}] = 1.34\\\\\rm pH &= \rm 14 - pOH\\\\\rm pH &= 12.66\end{aligned}[/tex]
Therefore, 12.66 is the pH of the mixture.
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My question today is : The reaction of 11.9g go CHCl3 with excess chlorine produced 10.2g of CCl4, carbon tetrachloride. What is the percent yield
Answer:
66.7%
Explanation:
Step 1: Write the balanced equation
CHCl₃ + Cl₂ ⇒ CCl₄ + HCl
Step 2: Calculate the theoretical yield of CCl₄ from 11.9 g of CHCl₃
According to the balanced equation, the mass ratio of CHCl₃ to CCl₄ is 119.38:153.82.
11.9 g CHCl₃ × 153.82 g CCl₄/119.38 g CHCl₃ = 15.3 g CCl₄
Step 3: Calculate the percent yield of CCl₄
Given the experimental yield of CCl₄ is 10.2 g, we can calculate the percent yield using the following expression.
%yield = (exp yield/theo yield) × 100%
%yield = (10.2 g/15.3 g) × 100% = 66.7%
How do I solve this?
Explanation:
a) Since this is a double displacement reaction, we write the balanced equation as
[tex]2AgNO_3(aq) + CaCl_2(aq) \\ \rightarrow 2AgCl(s) + Ca(NO_3)_2(aq)[/tex]
b) Next we find the number of moles of AgNO3 in the solution.
[tex](0.005\:\text{L})(0.500\:M\:AgNO_3) \\ = 0.0025\:\text{mol}\:AgNO_3[/tex]
Next, use the molar ratio to find the necessary amount of CaCl2 to react with the AgNO3:
[tex]0.0025\:\text{mol}\:AgNO_3× \left(\dfrac{1\:\text{mol}\:CaCl_2}{2\:\text{mol}\:AgNO_3} \right)[/tex]
[tex]= 0.00125\:\text{mol}\:CaCl_2[/tex]
The volume of 0.500 M solution of CaCl2 necessary to react all of the given AgNO_3 is then
[tex]V = \dfrac{0.00125\:\text{mol}\:CaCl_2}{0.500\:\text{M}\:CaCl_2}[/tex]
[tex]= 0.0025\:\text{L} = 2.5\:\text{mL}\:CaCl_2[/tex]
c) The theoretical yield can then be calculated as
[tex]0.0025\:\text{mol}\:AgNO_3 × \left(\dfrac{2\:\text{mol}\:AgCl}{2\:\text{mol}\:AgNO_3} \right)[/tex]
[tex]= 0.0025\:\text{mol}\:AgCl[/tex]
Converting this amount of AgCl into grams, we get
[tex]0.0025\:\text{mol}\:AgCl × \left(\dfrac{143.32\:\text{g}\:AgCl}{1\:\text{mol}\:AgCl} \right)[/tex]
[tex]= 0.358\:\text{g}\:AgCl[/tex]
Which list contains gases found only in "clean" air?
A. carbon dioxide, nitrogen dioxide, sulfur dioxide, helium
B. helium, methane, carbon dioxide, xenon
hydrogen, oxygen, chlorine, ammonia
C. argon, neon, krypton, radon
D. fluorine, chlorine, bromine, iodine
Answer:
hydrogen, oxygen, chlorine, ammonia
Explanation:
Air is a mixture of gases. When we say "clean" air here, we are referring to air that does not contain pollutant gases.
Some components of air such as water vapour, methane, CO2, and N2O are greenhouse gases. They are known to contribute towards global warming.
Some gases such as SO2 and NO2 contribute towards acid rain. The oxides of nitrogen are particularly involved in the formation of photochemical smog.
The halogens are known to lead to the depletion of the ozone layer and radon is a radioactive gas.
Hence, hydrogen, oxygen, chlorine, ammonia have no negative environmental impact hence they are found in clean air.
Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
Dato: Ce Cu: 0,093 cal/gºC; Ce Sn: 0,060 cal/gºC
URGENTEEEE
Answer:
La temperatura final del sistema es 1029,346 °C.
Explanation:
Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:
[tex]m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)[/tex]
[tex](m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}[/tex]
[tex]T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}[/tex] (1)
Donde:
[tex]m_{Sn}[/tex] - Masa del estaño, en gramos.
[tex]m_{Cu}[/tex] - Masa del cobre, en gramos.
[tex]c_{Sn}[/tex] - Calor específico del estaño, en calorías por gramo-grados Celsius.
[tex]c_{Cu}[/tex] - Calor específico del cobre, en calorías por gramo-grados Celsius.
[tex]T_{Sn}[/tex] - Temperatura inicial del estaño, en grados Celsius.
[tex]T_{Cu}[/tex] - Temperatura inicial del cobre, en grados Celsius.
Si sabemos que [tex]m_{Cu} = 150\,g[/tex], [tex]m_{Sn} = 35\,g[/tex], [tex]c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]T_{Sn} = 560\,^{\circ}C[/tex] y [tex]T_{Cu} = 1100\,^{\circ}C[/tex], entonces la temperatura final del sistema es:
[tex]T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}[/tex]
[tex]T = 1029,346\,^{\circ}C[/tex]
La temperatura final del sistema es 1029,346 °C.
Question 2: Functional Groups (6 points)
A. Methanoic acid is the simplest carboxylic acid molecule. It has one carbon atom. Draw the structural model for methanoic acid (using C and H). (3 points)
B. Methanal is the simplest aldehyde, with one carbon atom. Draw the structural model for methanal (using C and H). (3 points)
A student needed to make a 3 g/L NaCl solution. The student weighed 3 g of NaCl in a beaker and measured 1 L of water in a 1L volumetric flask that was labeled TC. The student then added the water to the beaker containing the NaCl. What errors did this student make? Describe how this can be performed properly.
Answer:
The answer is in the explanation.
Explanation:
A solution is defined as the homogeneous mixture of a solute (In this case, NaCl) and the solvent (water).
To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.
In a closed system, If a gas is transported to a container with double the volume of the previous container, the gas was held in, what is the gases' new volume?
The volume of the gas is fixed and will not change.
The volume of the gas will be half the original volume.
The volume of the gas will be the original volume squared.
The volume of the gas will be double the original volume.
Answer:
The volume of the gas is fixed and will not change.
Explanation:
The volume of the gas will not change because there is no change in temperature. Temperature increases the volume of gases enclosed in a container.
Someone can you please please help me
Answer:
FalseExplanation:
Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.
Cis-4-tertButylcyclohexyl bromide (compound 1) and Trans-4 tert Butylcyclohexylbromide (compound 2) are reacted with Potassium Tertiary butoxide in Tertiary butanol to produce 4-tertbutylcyclohexene. The following statement is completely true?A. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
B. In compound 1 the Tert butyl group occupies the axial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
C. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the equatorial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
D. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
E. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl occupies the equatorial and the bromine occupies axial position. Compound 1 reacts faster than compound 2.
F. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compond 2 reacts faster than compound 1.
Answer:
In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
Explanation:
In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.
Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.
This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.
