It matches the universal pH indicator and is indicating the proper pH
Did you intend to write [PdCl4]^-2 instead of PdCl2-4? If so, then:
<span>Cathode: [PdCl4]^-2(aq) + 2e- ======⇒ Pd(s) + 4Cl-(aq) </span>
<span>Anode: Cd(s) ==⇒ Cd+2(aq) + 2e-</span>
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
Answer: The molecular formula will be 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 70.6 g
Mass of H = 5.9 g
Mass of O = 23.5 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H =
For O =
The ratio of C : H: O= 4: 4:1
Hence the empirical formula is 
The empirical weight of = 4(12)+4(1)+1(16)= 68g.
The molecular weight = 136 g/mole
Now we have to calculate the molecular formula.
The molecular formula will be=
Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
