This is too hard to answer
Answer:
8.90
Explanation:
Density = mass ÷ volume
D = 222.50 g ÷ 25.00
= 8.9
The density of the unknown metal is 8.90.
Hope that helps.
<u>Answer:</u> The Gibbs free energy of the reaction is 21.32 kJ/mol
<u>Explanation:</u>
The chemical equation follows:
The equation used to Gibbs free energy of the reaction follows:
where,
= free energy of the reaction
= standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314J/K mol
T = Temperature = ![37^oC=[273+37]K=310K](https://tex.z-dn.net/?f=37%5EoC%3D%5B273%2B37%5DK%3D310K)
= Ratio of concentration of products and reactants = ![\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%5BOxaloacetate%5D%7D%5BNADH%5D%7D%7B%5Ctext%7B%5BMalate%5D%7D%5BNAD%5E%2B%5D%7D)
![\text{[Oxaloacetate]}=0.130mM](https://tex.z-dn.net/?f=%5Ctext%7B%5BOxaloacetate%5D%7D%3D0.130mM)
![[NADH]=2.0\times 10^2mM](https://tex.z-dn.net/?f=%5BNADH%5D%3D2.0%5Ctimes%2010%5E2mM)
![\text{[Malate]}=1.37mM](https://tex.z-dn.net/?f=%5Ctext%7B%5BMalate%5D%7D%3D1.37mM)
![[NAD^+]=490mM](https://tex.z-dn.net/?f=%5BNAD%5E%2B%5D%3D490mM)
Putting values in above expression, we get:
Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol
Answer:
Magnesium oxide
Explanation:
When magnesium reacts with oxygen, the magnesium atoms donate electrons to O2 molecules and thereby reduce the oxygen. Magnesium therefore acts as a reducing agent in this reaction. The O2 molecules, on the other hand, gain electrons from magnesium atoms and thereby oxidize the magnesium.
According to the kinetic theory, the mean free path is the average distance a single atom or molecule of an element or compound travels with respect with the other atoms during a collision. The greater the mean free path, the more ideal the behavior of a gas molecule is because intermolecular forces are minimum. To understand which factors affect the mean free path, the equation is written below.
l = μ/P * √(πkT/2m), where
l is the mean free path
μ is the viscosity of the fluid
P is the pressure
k is the Boltzmann's constant
T is the absolute temperature
m is the molar mass
So, here are the general effects of the factors on the mean free path:
Mean free path increases when:
1. The fluid is viscous (↑μ)
2. At low pressures (↓P)
3. At high temperatures (↑T)
4. Very light masses (↓m)
The opposite is also true for when the mean free path decreases. Factors that are not found here have little or no effect.
