<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer : The correct option is (3) 500 K and 0.1 atm.
Explanation :
A real gas behaves ideally at high temperature and low pressure.
The ideal gas equation is,

where,
P = pressure of gas
V = Volume of gas
R = Gas constant
T = temperature of gas
n = number of moles of gas
The ideal gas works properly when the inter-molecular interactions between the gas molecules and volume of gas molecule will be negligible. This is possible when pressure is low and temperature is high.
Therefore, the correct option is (3) 500 K and 0.1 atm.
Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.
<h3>Addition and decomposition reactions</h3>
Magnesium burns in air to produce magnesium oxide as follows:

Potassium manganate 7 burns to produce multiple products as follows:

Thus, the MgO will be heavier than Mg. On the other hand,
will be less heavy than
.
More on reactions can be found here: brainly.com/question/17434463
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The final temperature = 36 °C
<h3>Further explanation</h3>
The balanced combustion reaction for C₆H₆
2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ
MW C₆H₆ : 78.11 g/mol
mol C₆H₆ :

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

Heat transferred to water :
Q=m.c.ΔT

Answer: 4
Explanation:
Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.
Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as
. The value of this quantum number ranges from
. When l = 2, the value of
will be -2, -1, 0, +1, +2.
Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.
Also electrons give are 32.
The formula for number of electrons is
.


Thus principal quantum no will be n= 4.