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Answer:
Explanation: k =10^3, m =10^-3
2.8 Kg = 2.8 *10^3 g
To conver kg to mg you need to
=2.8* 10^3 *10^-3* 10^3
=2.8 * 10^6 *10^-3
= 2.8 * 10^6 mg
Answer:
545
Explanation:
If we let the unknown temperature in Kelvin be x,
4.5 × 10⁻⁵ L of NH₃ are needed per liter of flue gas at 1.00 atm.
<h3>What is Balanced Chemical Equation ?</h3>
The equation during which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation is called balanced chemical equation.
The given balanced chemical equation is
4NH₃(g) + 4NO(g) + O₂(g) → 4N₂(g) + 6H₂O(g)
<h3>How to find mole fraction ?</h3>
To find mole fraction use this expression
P.P = M.F × T.P
where,
P.P = Partial Pressure
M.F = Mole Fraction
T.P = Total pressure
Here,
Given flue gas total pressure (T.P) is 1.00 atm
Given NO Partial Pressure (P.P) = 4.5 × 10⁻⁵
Now putting the values we get
P.P = M.F × T.P
4.5 × 10⁻⁵ atm = 1.00 atm × M.F
M.F = 4.5 × 10⁻⁵ L
Thus from the above conclusion we can say that 4.5 × 10⁻⁵ L of NH₃ are needed per liter of flue gas at 1.00 atm.
Learn more about Mole Fraction here: brainly.com/question/1601411
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Answer:
0.4076 g
Explanation:
Kp is the equilibrium constant based on pressure and depends only on gas substances. For a generic reaction
aA + bB ⇄ cC + dD
, where pX is the pressure of X in equilibrium.
For the reaction Kp = pCO₂
pCO₂ = 0.026 atm
The system is in equilibrium at the beginning. The compression occurs at a constant temperature, so using Boyle's law
P1V1 = P2V2
0.026*10 = P2*0.1
P2 = 2.6 atm
The reaction will reach again the equilibrium, and pCO₂ = 0.026 atm, then the rest will form MgCO₃, which will be 2.6 - 0.026 = 2.574 atm.
By the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature.
2.574*0.1 = n*0.082*650
53.3n = 0.2574
n = 4.83x10⁻³ mol
The stoichiometry of the reaction is 1 mol of MgCO₃ for 1 mol of CO₂, so it will form 4.83x10⁻³ mol of MgCO₃ .
The molar mass is:
MgCO₃: 24 g/mol of Mg + 12 g/mol of C + 3*16 g/mol of O = 84 g/mol
The mass formed is the molar mass multiplied by the number of moles:
m = 84x4.83x10⁻³
m = 0.4076 g