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Maslowich
2 years ago
14

A 15.8 g sample contains 3.60 g F, 4.90 g H, and 7.30 g C. What is the percent composition of hydrogen in this sample?

Chemistry
1 answer:
deff fn [24]2 years ago
3 0

Answer:

The correct answer is "32%".

Explanation:

The given values:

Weight of H,

= 4.9 g

Weight of sample,

= 15.8 g

Now,

The weight percentage of C will be:

= \frac{Weight \ of \ C}{Total \ weight}\times 100

By substituting the values, we get

= \frac{4.9}{15.8}\times 100

= 32 \ percent

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Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
Mass in grams of 6.25 mol of copper (II) nitrate?
podryga [215]
Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997

Copper (II) Nitrate has a MW of 125.544 g/mol

6.25 x 125.544

= 784.65 <--- is your answer, if there were was a multiple choice or not :)
8 0
2 years ago
Read 2 more answers
analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula
MrMuchimi
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr   
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr  /0.0135= 1 mol Cr
 0.0538 mol O/0.035= 4 mol Cr
K2CrO4
6 0
2 years ago
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water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water
Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0
2 years ago
Fill in the blanks with the correct word.
Vitek1552 [10]

Answer:

1. During diffusion, when the concentration of molecules on both sides of a membrane is the same, the molecules will continue to move across the membrane in both directions.

Hopes it Helps!

6 0
2 years ago
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