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After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml
<h3>Subtraction of Numbers</h3>
Given Data
- Volume of Hexane dispensed = 40ml
Let us compute the amount of excess hexane/ the volume that will remain
Remainder = The difference in volume dispensed and the volume needed
Remainder = 40-5
Remainder = 35 ml
The remainder is 35ml
Learn more about subtraction of numbers here:
brainly.com/question/4721701
Answer:
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Explanation:
The balanced reaction between Na2CO3 and HCl is given as;
Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)
The next step is o express the species as ions.
The complete ionic equation for the above reaction would be;
2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)
The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.
The spectator ions are; Na⁺ and Cl⁻
The net ionic equation is given as;
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Answer:
molar mass M(s) = 65.326 g/mol
Explanation:
- M(s) + H2SO4(aq) → MSO4(aq) + H2(g)
∴ VH2(g) = 231 mL = 0.231 L
∴ P atm = 1.0079 bar
∴ PvH2O(25°C) = 0.03167 bar
Graham´s law:
⇒ PH2(g) = P atm - PvH2O(25°C)
⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm
∴ nH2(g) = PV/RT
⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))
⇒ nH2(g) = 9.1082 E-3 mol
⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))
⇒ n M(s) = 9.1082 E-3 mol
∴ molar mass M(s) [=] g/mol
⇒ molar mass M(s) = (0.595 g) / (9.1082 E-3 mol)
⇒ molar mass M(s) = 65.326 g/mol
* 184 gEther/mol = 0.7028 g