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3 years ago

When zinc or aluminum was allowed to react with the copper sulfate , what was the limiting reagen?

1 answer:

5 0

Usually it is the CuSO4 that is the limiting reagent.

if all of the color of the solution was gone, but there was still some zinc metal mixed in with the copper metal produced, then Zn is the excess reagent 

f all of the color of the solution was not gone, but there was no zinc metal left in with the blue copper solution , then Zn is the limiting reagent Hope this helps.

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Which of the substances shown below most likely exhibit hydrogen bonding with other identical

navik [9.2K]

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2 years ago

Read 2 more answers

Ppredict the identity of the precipitate in the below reaction: BaCl2(aq) + K2SO4(aq) →

Annette [7]

Answer: The precipitate formed is $BaSO_4$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid or precipitated form are represented by (s) after their chemical formulas.

A double displacement reaction in which one of the product is formed as a solid is called as precipitation reaction.

The balanced chemical equation is:

$K_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2KCl(aq)$

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2 years ago

Describe how visible spectroscopy could be used to determine the order of a reaction in which a single colorless reactant produc

d1i1m1o1n [39]

Answer:

Explanation:

In a reaction, where, one of the reactant produces a colored product, visible spectroscopy can be used to determined the order of a reaction, the change in concentration of the reactant which forms the colored product is determined by absorbance measurement over time. The data for the concentration and time are plotted on the y and x axis and If we get a straight line it is a zero-order reaction. If instead, a plot of ln[concentration] versus time gives a straight line, it is a first order reaction. However, If 1/concentration versus time gives a straight line, it is a second order reaction kinetics. The other reactants may be changed while keeping this reactant as constant and change on rate of the reaction is observed to see If the other reactant affects the reaction or not.

8 0

3 years ago

A rigid container of gas has a pressure of 1.72 atm and temperature of 21 oC. If the temperature increases to 85 oC, what is the

Vinil7 [7]

Answer:

the new pressure is 2.09 atm

Explanation:

you have to use gay lussac's law so the formula is

p1/t1 = p2/t2

and convert C to Kelvin k=C+273.15

1.72atm/294.15 = p2/358.15

solve for p2 by multiplying 358.15 on both sides

p2=2.09 atm

5 0

2 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

2 years ago