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nadezda [96]
3 years ago
7

AdiffusioncoupleofweldedAandBwasgivenadiffusionannealatelevatedtemperatureandthencooledtoroomtemperature.Chemicalanalysisofthesu

ccessivelayercutparalleltotheweldinterfaceshowthatoveradistanceof5000nm,NAchangedfrom0.3to0.35.Assumethatthenumberofatomsperm3ofbothpuremetalsis9×1028.a)determinetheconcentrationgradientdnA/dx,b)ifthediffusioncoefficientatthepointinquestionatannealingtemperaturewas2×10-14m2/s,determinethenumberofAatomspersecondthatwouldpassthroughthiscross-sectionattheannealingtemperature.Neglecttheinterfacemotion.
Chemistry
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

what

Explanation:

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Precipitation reactions always occur when two aqueous solutions are mixed (T/F)
Studentka2010 [4]

Answer: The given statement is false.

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Here AgCl is present in solid state so, it is the precipitate.

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For example, NaCl(aq) + KNO_{3}(aq) \rightarrow KCl(aq) + NaNO_{3}(aq)

Hence, we can conclude that the statement precipitation reactions always occur when two aqueous solutions are mixed, is false.

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Put the following steps of specimen preparation and staining in order: I. Application of staining dyes II. Heat fixation III. Sm
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II

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3 years ago
The chief advantage of the metric system over other systems of measurement is that it ____.
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7 0
2 years ago
Read 2 more answers
A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f
Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0
2 years ago
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