Answer:
Iron is limiting reactant
Explanation:
Based on the reaction:
Fe + S → FeS
1 mole of iron reacts per mole of Sulfur
To solve this question we must convert the mass of each reactant to moles using molar masses of each reactant. As the reaction is 1:1, the reactant with the lower amount of moles is limiting reactant.
Moles Fe -Molar mass: 55.845g/mol-
4.68g * (1mol / 55.845g) = 0.0838 moles
Moles S -Molar mass: 32.065g/mol-
2.88g * (1mol / 32.065g) = 0.0898 moles
As the amount of moles of Fe < Moles S,
Iron is limiting reactant
When 4.68 g of iron reacts with 2.88 g of sulfur to produce FeS, iron is the limiting reagent.
What is limiting reagent?
If in a chemical reaction two reactants are present and one of them is present in less quantity as compared to other, is known as limiting reagent.
Given chemical reaction is:
Fe + S → FeS
From the stoichiometry of the reaction it is clear that equal moles of both reactant is required for the formation of product, so their mole ratio is 1:1.
Now we calculate the moles by using the formula:
n = W/M, where
W = given mass
M = molar mass
Moles of 4.68g of iron = 4.68g / 55.845g/mole = 0.0838 moles
Moles of 2.88 of sulfur = 2.88g / 32.065g/mole = 0.0898 moles
Moles of iron is less as compare to the sulfur, so it is the limiting reagent.
Hence, iron is the limiting reagent.
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5). At what temperature (K) will 0.854 moles of neon gas occupy 12.3 L at 1.95
atmospheres?
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
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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'.
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.
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(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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Meera added blue copper sulphate crystals to some water in a beaker.
The copper sulphate dissolved in the water.
1 give one way meera could see that the copper sulphate had dissolved in the
Answer:
The solid crystals disappeared
Explanation:
When a soluble solid solute is added to water, the solid solute disappears after a little while. The disappearance of this solute indicates that the solute has been dissolved in water.
In this case, blue copper sulphate crystals are added to water, the blue crystals disappear leaving only a blue solution. The disappearance of these blue copper sulphate crystals indicates that the substance has dissolved in water.
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
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
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: