Answer:
338.38 K
Explanation:
Applying,
PV = nRT............... Equation 1
Where P = pressure, V = Volume, R = Temperature, n = number of moles, T = temperature.
Make T the subject of the equation,
T = PV/nR............. Equation 2
From the question,
Given: P = 1.95 atm, V = 12.3 L, n = 0.854 moles
Constant: R = 0.083 L.atm/K.mol
Substitute these values into equation 2
T = (1.95×12.3)/(0.854×0.083)
T = 338.38 K
(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction, how much copper(II) nitrate is also produced? Answer in units of mol. (part 2 of 3) How much Cu is required in this reaction? Answer in units of mol. (part 3 of 3) 1.0 points How much AgNO3 is required in this reaction? Answer in units of mol.
Answer:
See explanation.
Explanation:
Hello there!
In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:
[tex]Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2[/tex]
Thus, we proceed as follows:
Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:
[tex]m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2[/tex]
Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:
[tex]m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu[/tex]
Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:
[tex]m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3} \\\\m_{AgNO_3}=2.03gAgNO_3[/tex]
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A chemist observed bubbling and fizzing after adding an acid solution to a
white powdery substance in a beaker. Which of the following can be
inferred?
Answer:
a chemical reaction occured
Explanation:
bubbling and fizzing after adding a substance, most offten means a chemical reaction is happening
When writing the formulas for a compound that contains a polyatomic ion, ... ?
Answer:
The cation is written first in the name; the anion is written second in the name. Rule 2. When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
When writing the formula of a compound that contains polyatomic ion, the metal is written first followed by the central atom in the ion and then other atoms that surround the central atom.
A poly atomic ion refers to an ion that comprises of more than one atom. Such ions are common in chemistry. Examples of polyatomic ions include; PO4^3-, BH4^- etc.
When writing the formula of a compound that contains a polyatomic ion, the metal is written first then the central atom in the ion follows before other atoms that surround the central atom in the ion.
Learn more: https://brainly.com/question/6284546
Help calculate the density and please type it like this and explain how you got the answer thanks
Mass:
Volume;
Density:
Answer: 2.5
Explanation:
density= 2.5
mass= 62.5 or 62.50
volume= 25 ( 75-50= 25)
so divide 62.5( or 62.50) by 25 which will led you to 2.5
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(11) Deduce the number of dyes in food colouring H.
(iii) Suggest why food colouring F does not move during the experiment.
(iv) Explain which two food colourings contain the dye that is likely to be the most soluble the solvent.
(b) Determine which food colouring contains a dye with R, value closest to 0.67
Show your working.
Answer:
(ii) 1 dye
(iii) Food coloring F is insoluble in the solvent
(iv) 'E' and 'H'
(b) Food colouring G
Explanation:
Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper
The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper
(ii) The number of components into which the food colouring 'H' separates into = 1
Therefore, the number of dyes in food colouring 'H' = 1 dye
(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase
(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;
Food coloring 'E' and 'H'
(b) Using a similar question solution found on 'tutor my self' website, we have;
The [tex]R_f[/tex] values are given as follows;
[tex]R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}[/tex]
The distance moved by the solvent = 5 units
The distance moved by dyes in food colouring 'E' and 'H' = 4 units
The distance moved by dye in food colouring 'G' = 3.3 units
The distance moved by the second dye in food colouring 'E' = 2.7 units
By inspection, we get;
[tex]R_f[/tex] dye in food colouring 'G' = 3.3/5 = 0.66,
Therefore, the dye with [tex]R_f[/tex] value closest to 0.67 is the dye in food colouring 'G'.
Use the given Nernst equation and reaction to solve this problem. What is the potential of this cell with the given conditions?
2Li (aq) + F2(g) 2Li+(aq) + 2F- (aq)
E° = +5.92 volts
T = 200°C
[Li+] = 10.0 molar
[F-] = 10.0 molar
Answer:
The 2nd one is the one
Explanation:
and it isn't writen out all the way
why do we need to rinse the mouth before collecting the saliva
What are the two limitations of earth plates
Answer:
The tectonic style and viability of modern plate tectonics in the early Earth is still debated. Field observations and theoretical arguments both in favor and against the uniformitarian view of plate tectonics back until the Archean continue to accumulate. Here, we present the first numerical modeling results that address for a hotter Earth the viability of subduction, one of the main requirements for plate tectonics. A hotter mantle has mainly two effects: 1) viscosity is lower, and 2) more melt is produced, which in a plate tectonic setting will lead to a thicker oceanic crust and harzburgite layer. Although compositional buoyancy resulting from these thick crust and harzburgite might be a serious limitation for subduction initiation, our modeling results show that eclogitization significantly relaxes this limitation for a developed, ongoing subduction process. Furthermore, the lower viscosity leads to more frequent slab breakoff, and sometimes to crustal separation from the mantle lithosphere. Unlike earlier propositions, not compositional buoyancy considerations, but this lithospheric weakness could be the principle limitation to the viability of plate tectonics in a hotter Earth. These results suggest a new explanation for the absence of ultrahigh-pressure metamorphism (UHPM) and blueschists in most of the Precambrian: early slabs were not too buoyant, but too weak to provide a mechanism for UHPM and exhumation.
Explanation: