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sashaice [31]
3 years ago
10

The half-reaction occurring at the anode in the balanced reaction shown below is ________. 3mno4- (aq) + 24h+ (aq) + 5fe (s) → 3

mn2+ (aq) + 5fe3+ (aq) + 12h2o (l)
Chemistry
1 answer:
alexgriva [62]3 years ago
3 0
Reduction takes place in cathode while oxidation takes place at the anode.
Given that the reaction 3MnO4- +24H + +5Fe→3Mn+2+5Fe+3+12H2O
Now that oxidation is termed as an increase in oxidation number or loss of electrons while reduction is a decrease in oxidation number and gain of electrons,
∴oxidation will be
Fe→Fe +3+3e-
reduction will
MnO4- +8H+ →Mn+2+4H2O
If oxidation takes place at anode then Anode: an oxidation reaction
Fe→Fe+3+3e- . Then the answer is 
Fe(s)→Fe+3(aq)+3e-
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A 2.00 L flask is filled with propane gas (C3H8) at a pressure of 1.10 atm and a temperature of 258 K. What is the mass of propa
pychu [463]

Answer:

m = 4.58 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this problem is solved via the ideal gas equation:

PV=nRT

As we can calculate the moles of propane given the pressure, temperature and volume as shown below:

n=\frac{PV}{RT} \\\\n=\frac{1.10atm*2.00L}{0.08206\frac{atm*L}{mol*K}*258K} \\\\n=0.104mol

Finally, since the molar mass of propane is 44.1 g/mol, we calculate the mass by following the shown below mole-mass conversion factor:

m=0.104mol*\frac{44.1g}{1mol}\\\\m= 4.58g

Regards!

3 0
3 years ago
Radiation that changes direction as it passes through material is called what
RSB [31]
Light? Look up the Third law. 
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7 0
3 years ago
A sample of gas is held at 100oc at a volume of 20 L.if the volume is increased to 40 L what is the new temperature of the gas i
ElenaW [278]

Answer:

470 °C  

Explanation:

This looks like a case where we can use Charles’ Law:  

\dfrac{V_{1}}{T_{1}} =\dfrac{V_{2}}{T_{2}}

Data:

V₁ = 20 L; T₁ = 100 °C

V₂ = 40 L; T₂ = ?  

Calculations:

(a) Convert the temperature to kelvins

T₁ = (100 + 273.15) K = 373.15 K

(b) Calculate the new temperature

\begin{array}{rcl}\dfrac{V_{1}}{T_{1}}& =&\dfrac{V_{2}}{T_{2}}\\\\ \dfrac{\text{20 L}}{\text{373.15 K}} &=&\dfrac{\text{40 L}}{T_{2}}\\\\{\text{15 000 K}} & = & 20T_{2}\\T_{2} & = &\dfrac{\text{15 000 K}}{20 }\\\\T_{2} & = & \textbf{750 K}\\\end{array}

Note: The answer can have only two significant figures because that is all you gave for the volumes.

(c) Convert the temperature to Celsius

T₂ = (750 – 273.15) °C = 470 °C

3 0
3 years ago
Can u help?? It's science and I'm in 6th grade
kvv77 [185]
Yeah I can help I'm an A student in science
6 0
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