5). At what temperature (K) will 0.854 moles of neon gas occupy 12.3 L at 1.95
atmospheres?

Answer:

See explanation.

Explanation:

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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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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.

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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

Answer:

a chemical reaction occured

Explanation:

bubbling and fizzing after adding a substance, most offten means a chemical reaction is happening

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'.

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:

Answer:

The 2nd one is the one

Explanation:

and it isn't writen out all the way