Calculate the mass of 1.00 mol of each of these substances

vichka [17]

Answer:

2Na+F2 yields 2NaF is balanced.

Explanation:

There are 2 sodium and 2 fluorine in both reactants and product: In 2NaF the 2 is distributed because it is in the beginning of the compound.

7 0

3 years ago

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is carried out adiabatically in a constant-volume batch reactor. The rate law is Plot and analyze the conversion, temperature, a

Sphinxa [80]

Answer:

See explaination

Explanation:

The mole balance for a constant-volume batch reactor is given such as, For a first-order isothermal reaction, the time to reach a given conversion is the same for constant-pressure and constant-volume reactors. Also, the time is the same for a reaction of any order if there is no change in the number of moles.

Please kindly check attachment for the step by step solution of the given problem.

5 0

3 years ago

An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t

erik [133]

Answer:

$\% Fe^{+2}=70%$

Explanation:

Hello,

In this case, we could considering this as a redox titration:

$Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Thus, the balance turns out (by adding both hydrogen ions and water):

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-$

Thus, by stoichiometry, the grams of Fe+2 ions result:

$m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}$

Finally, the mass percent is:

$\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%$

Best regards.

8 0

3 years ago

A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom

Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

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polyalchemVirtuoso

Answer:

Explanation:

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34kurt

Answer: The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.

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