Answer:
The temperature of the gas is 876.69 Kelvin
Explanation:
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
In this case:
- P= 470 mmHg
- V= 570 mL= 0.570 L
- n= 0.216 g= 0.0049 moles (being the molar mass of carbon dioxide is 44 g/mole)
- R= 62.36367
Replacing:
470 mmHg*0.570 L= 0.0049 moles* 62.36367 *T
Solving:
T= 876.69 K
<em><u>The temperature of the gas is 876.69 Kelvin</u></em>
<span>Which reagent will be used up first? Thee answer is HCl.
We need to convert the amounts of the reactants into moles to see the limiting reactant.
48.2 g HCl ( 1 mol / 36.46 g) = 1.32 mol HCl
Since the ration is 1:4 then the limiting reactant is HCl. To react all of the MnO2 we need 3.44 mol HCl.
</span><span>How many grams of Cl2 will be produced?
</span>We use the amount of limiting reactant since it will be used up first.
1.32 mol HCl (1 mol Cl2 / 4 mol HCl) ( 70.9 g Cl2 / 1 mol Cl2) =23.4 g CL2
Molecular weight= 1.00784 u
mass of one mole = 1.008 grams
Answer
MnO₄ + 2H⁺ +3NO₂⁻ →3NO₃⁻ + Mn²⁺ +H₂O
Explanation
This is a redox reaction (oxidation-reduction reaction) which involves the transfer of electrons between two species. i.e
Mn + 6e⁻→Mn²⁺ (reduction)
3N³⁺- 6e⁻→3Mn⁵⁺(oxidation)
The phenomenon known as "salting-out" occurs at very high ionic strengths, when protein solubility declines as ionic strength rises. As a result, salting out may be used to segregate proteins according to how soluble they are in salt solutions.
Because large levels of sodium chloride disturb the bonds and structure of the active site, the rate of enzyme activity will gradually decrease as the concentration of sodium chloride rises. As a result, some of the active sites get denaturized and the starch loses its ability to attach to them. As more enzymes get denatured and eventually cease to function, enzyme activity will steadily wane.
