Answer:
c) 22
Explanation:
Let's consider the following balanced equation.
N₂(g) + 3 H₂(g) ----> 2 NH₃(l)
According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:
According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:
The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.
We can calculate the pressure (P) using the ideal gas equation.
P.V = n.R.T
where
V is the volume (50.0 L)
n is the number of moles (20.0 mol)
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (400.0 + 273.15 = 673.2K)
Answer: 91.73g of NaCl
Explanation:
First, we solve for the number of moles of F2 using the ideal gas equation
V = 12L
P = 1.5 atm
T = 280K
R = 0.082atm.L/mol/K
n =?
PV = nRT
n = PV /RT
n = (1.5x12)/(0.082x280)
n = 0.784mol
Next, we convert this mole ( i.e 0.784mol) of F2 to mass
MM of F2 = 19x2 = 38g/mol
Mass conc of F2 = n x MM
= 0.784 x 38 = 29.792g
Equation for the reaction is given below
F2 + 2NaCl —> 2NaF + Cl2
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass conc. of NaCl from the equation = 2 x 58.5 = 117g
Next, we find the mass of NaCl that reacted with 29.792g of F2.
From the equation,
38g of F2 redacted with 117g of NaCl.
Therefore, 29.792g of F2 will react with Xg of NaCl i.e
Xg of NaCl = (29.792 x 117)/38
= 91.73g
Therefore, 91.73g of NaCl reacted with f2
Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero,
for NiO4-, assuming oxidation number of Ni is x
(x ×1) + (-2 × 4) = -1
x = + 7
Therefore, the oxidation number goes from 0 to +7
Answer:
The 2 zeros after the decimal point
Answer:
38503.5N
Explanation:
Data obtained from the question include:
P (pressure) = 5.00 atm
Now, we need to convert 5atm to a number in N/m2 in order to obtain the desired result of force in Newton (N). This is illustrated below:
1 atm = 101325N/m2
5 atm = 5 x 101325 = 506625N/m^2
A (area of piston) = 0.0760 m^2
Pressure is force per unit area. Mathematically it is written as
P = F/A
F = P x A
F = 506625 x 0.0760
F = 38503.5N
Therefore, the force exerted on the piston is 38503.5N
