<span>0.967 x 10^-6 HZ
This should be correct (:</span>
First, we need to get moles of NaOH:
when moles NaOH = volume * molarity
= 0.02573L * 0.11 M
= 0.0028 moles
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
= 9.4 x 10^-4 / 0.034 L
= 0.028 M
202827.0000 is the answer I think idrk
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
