(2) gains 12 moles of electrons, easy, because it must retain the same mass in a balanced equation, so if there is no exterior mention of any electrons, such as the notation +12e-, than we must assume that the Oxygen took it at any rate.
Answer:
glcolysis
Explanation:
Glcolysis uses glucose to create ATP within the cytoplasm of the cell.
<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
From the balance and the growth you are talking to me, we can say that <span>it would be an inequality of 2x > 4x +12 so in that case it would be in the motnh number 6 where you would find a greater balance.</span>
Answer:
d =~ 5.8μm
d =~ 0.13 μm
Explanation:
when the doping concentrations are 5 × 10^15 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^15
d = 1/ 170997.5
d = 5.85 × 10 ^ -6
d =~ 5.8μm
when the doping concentrations are 5 × 10^20 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^20
using the principle of surds and standard forms, we have
d = 1/ ∛0.5 × 10^21
d = 1/7937005.26
d = 1.26 × 10 ^ -7
d = 0.126 × 10 ^ -6
d =~ 0.13 μm
