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Olenka [21]
3 years ago
8

What is the coefficient of mno2 when you balance the equation for this redox reaction? kcn(aq) + kmno4(aq) + h2o(l) → kcno(aq) +

mno2(s) + koh(aq)?
Chemistry
2 answers:
Tasya [4]3 years ago
7 0
Kcn(aq)+kmno4(aq)+h2o(l)----->2mno2(s)+koh(aq)

MariettaO [177]3 years ago
4 0
Taking a test with this question on it. I believe the answer is 2
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an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem
OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

P\cdot V = n\cdot R\cdot T,

\displaystyle V = \frac{n\cdot R\cdot T}{P}.

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of moles of particles in this gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of R will be 8.314 if P, V, and T are in SI units. Convert these values to SI units:

  • P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa;
  • V shall be in cubic meters, \rm m^{3};
  • T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K.

Apply the ideal gas law:

\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}.

4 0
3 years ago
This reaction appears to be destroying the logs
Alex_Xolod [135]
What exactly is the question you are asking?
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3 years ago
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