The citation you provided is from a scientific article titled "Large-Area Single-Layer MoSe2 and Its Van der Waals Heterostructures" published in ACS Nano in 2014 by Shim, G. W. and colleagues. The article discusses the synthesis and properties of single-layer MoSe2 and its van der Waals heterostructures.
MoSe2 is a material made up of molybdenum and selenium atoms arranged in a two-dimensional lattice. The article focuses on the production of large-area single-layer MoSe2, which refers to a single layer of atoms stacked on top of each other. This is significant because the properties of materials can change when they are in a two-dimensional form.
The researchers also explore van der Waals heterostructures, which are created by stacking different two-dimensional materials on top of each other. These heterostructures can exhibit unique properties that are different from the individual materials alone. For example, the electrical, optical, and mechanical properties of the heterostructure may be different from those of the individual layers.
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at constant temperature, a 144.0 ml sample of gas in a piston chamber has a pressure of 2.25 atm. calculate the pressure of the gas if this piston is pushed down hard so that the gas now has a volume of 36.0 ml.
The pressure of the gas would be 9.0 atm if the piston is pushed down hard to a volume of 36.0 ml.
To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume at constant temperature.
First, we need to set up the equation: P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Given that the initial volume (V1) is 144.0 ml and the initial pressure (P1) is 2.25 atm, and the final volume (V2) is 36.0 ml, we can plug in the values into the equation:
2.25 atm * 144.0 ml = P2 * 36.0 ml
Next, we can solve for P2 by dividing both sides of the equation by 36.0 ml:
2.25 atm * 144.0 ml / 36.0 ml = P2
P2 = 9.0 atm
Therefore, the pressure of the gas would be 9.0 atm if the piston is pushed down hard to a volume of 36.0 ml.
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A graduated cylinder contains 26 cm3 of water. an object with a mass of 21 grams and a volume of 15 cm3 is lowered into the water. what will the new water level be
When the object with a volume of 15 cm3 is lowered into the water in the graduated cylinder, the new water level will be 11 cm3.
The new water level in the graduated cylinder can be determined by considering the principle of displacement. When the object is lowered into the water, it will displace an amount of water equal to its own volume.
Given that the object has a volume of 15 cm3, it will displace 15 cm3 of water. Since the initial volume of water in the graduated cylinder is 26 cm3, the new water level can be calculated by subtracting the volume of water displaced by the object from the initial volume of water.
Therefore, the new water level in the graduated cylinder will be 26 cm3 - 15 cm3 = 11 cm3.
To summarize, when the object with a volume of 15 cm3 is lowered into the water in the graduated cylinder, the new water level will be 11 cm3.
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The liquid dispensed from a burette is called ___________.
i. solute
ii. water
iii. titrant
iv. analyte
The liquid dispensed from a burette is called the titrant. A titrant is a solution with a known concentration that is added in a controlled manner to react with the analyte in a chemical analysis. The option C is correct.
The burette is a precise measuring instrument used in titrations to deliver the titrant.In a titration, the analyte is the substance being analyzed or tested. It reacts with the titrant to form a product, and the reaction is monitored to determine the concentration or amount of the analyte.
For example, in an acid-base titration, a solution of known concentration called the titrant is slowly added to the analyte solution until the reaction between the acid and base is complete. The burette allows for precise measurement of the volume of titrant added.The other options given are not accurate in this context. Solute refers to the substance being dissolved in a solvent, while water is a common solvent. Analyte, as mentioned earlier, is the substance being analyzed. The correct term for the liquid dispensed from a burette in a titration is the titrant.
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How will the line techniqur differ when using a michanical pencil as compered to using an automatic pencil?
The line technique may differ between a mechanical pencil and an automatic pencil in terms of lead thickness, consistency, mechanism, and ergonomics, affecting line width, control, and user comfort.
The line technique may differ when using a mechanical pencil compared to an automatic pencil due to several factors:
Lead Thickness: Mechanical pencils come with various lead thickness options (e.g., 0.5mm, 0.7mm, etc.), while automatic pencils typically have a fixed lead size. The lead thickness affects the line's width, with thinner leads producing finer lines.
Consistency: Automatic pencils usually offer a constant lead length, resulting in a consistent line width. Mechanical pencils might require periodic advancement of the lead, which could lead to variations in line thickness if not adjusted uniformly.
Mechanism: Mechanical pencils employ a mechanical push mechanism, while automatic pencils utilize gravity or button press to advance the lead. This mechanical difference might influence the smoothness and control of the lines drawn.
Ergonomics: The design and grip of mechanical pencils may differ from automatic pencils, affecting the user's comfort and stability while drawing lines.
Overall, both pencil types can produce precise lines, but the line technique might vary in terms of thickness, consistency, and ease of use based on the specific pencil design and lead advancement mechanism.
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a homogeneous solution contains copper(ii) ions (cu2 ), silver ions (ag ) and potassium ions (k ). you have sodium bromide (nabr) and sodium sulfide (na2s) available to use. what should you add and in what order to separate the three metal ions? ksp (sulfides) ksp (bromides) cus 6.0×10–37 cubr2 soluble ag2s 6.0×10–51 agbr 7.7×10–13 k2s soluble kbr soluble
To separate Cu2+, Ag+, and K+ from the homogeneous solution, add sodium sulfide (Na2S) first to precipitate CuS. Then add sodium bromide (NaBr) to precipitate AgBr. Finally, the remaining solution contains only K+.
To separate the copper (II), silver, and potassium ions from the homogeneous solution, you can employ the following procedure.
Firstly, add sodium sulfide (Na2S) to the solution, resulting in the formation of insoluble copper sulfide (CuS) precipitate due to its low solubility (Ksp = 6.0×10–37). By filtering the solution, the insoluble CuS precipitate can be separated.
Next, introduce sodium bromide (NaBr) to the filtrate, causing the formation of insoluble silver bromide (AgBr) precipitate due to its low solubility (Ksp = 7.7×10–13). By filtering the solution once again, the insoluble AgBr precipitate can be isolated.
Finally, the remaining solution will only contain potassium ions (K+), which do not require further separation steps as potassium salts are highly soluble in water. By following this procedure, effective separation of the copper (II), silver, and potassium ions can be achieved.
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You prepare a stock solution that has a concentration of 2. 5 m. An aliquot with a volume of 10. 0 ml is removed from the solution. What is the concentration of the aliquot?.
The concentration of the aliquot is 2.5 M.
The concentration of a solution is defined as the amount of solute present per unit volume of the solution.
In this case, the stock solution has a concentration of 2.5 M (moles per liter).
An aliquot is a small portion or sample taken from a larger solution. In this scenario, an aliquot with a volume of 10.0 ml is removed from the stock solution.
Since the concentration of the stock solution is given in terms of moles per liter (M), the concentration of the aliquot will be the same as the concentration of the stock solution.
The concentration does not change when a specific volume is removed from the solution.
Therefore, the concentration of the aliquot is 2.5 M. It is important to note that the concentration remains the same regardless of the volume of the aliquot, as long as the proportion of solute to solvent remains constant.
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