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3 years ago

9

SOMEONE PLEASE HELP HURRY

1 answer:

melomori [17]3 years ago

4 0

I think the answer is A. I know it's Planet X.

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A potassium bromide solution is 8.34% potassium bromide by mass and its density is 1.03 g/ml. what mass of potassium bromide is

jenyasd209 [6]

Answer:- 3.84 grams

Solution:- Volume of the sample is 44.8 mL and the density is 1.03 gram per mL.

From the density and volume we calculate the mass as:

mass = volume*density

$44.8mL(\frac{1.03g}{mL})$

= 46.1 g

From given info, potassium bromide solution is 8.34% potassium bromide by mass. It means if we have 100 grams of the solution then 8.34 grams of potassium bromide is present in. We need to calculate how many grams of potassium bromide are present in 46.1 grams of the solution.

The calculations could easily be done using dimensional analysis as:

$46.1gSolution(\frac{8.34gKBr}{100gSolution})$

= 3.84 g KBr

Hence, 3.84 grams of KBr are present in 44.8 mL of the solution.

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3 years ago

The atomic number tells you how many ____ are in the ____ of an atom of that element

GrogVix [38]

<span>The atomic number tells you how many _PROTONS_ are in the _NUCLEUS_ of an atom of that element. :)</span>

6 0

3 years ago

Explain theÂ·relationship among atoms, elements, and compounds.

Minchanka [31]

DFBHVBHGHFGSDGBN CVXBFGHRTSHBTGF

3 0

3 years ago

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

<u>Reduction half reaction:</u> $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

<u>Net cell reaction:</u> $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

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:

$E^o_{cell}=0.799-(-0.762)=1.561V$

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

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

Can a chemical reaction be both exothermic and endothermic? Explain.

Elis [28]

Explanation:

reaction cannot be both endothermic and exothermic at the same time.

8 0

3 years ago