Chemical element of atomic number 22, a hard silver-gray metal of the transition series, used in strong, light, corrosion-resistant alloys.<span>These alloys are mainly </span>used<span> in aircraft, spacecraft and missiles because of their low density and ability to withstand extremes of temperature.</span>
Naming conventions for 2 non-metals like Si and O are based on their valence electrons, Si has 4 electrons around it and Oxygen has 6, in order for you to satisfy octet (8 electrons around each element) surrounding each Si and O, you need another O, To name these 2, just write the name of the first element which has less electrons first then the second element to which you use a prefix "di" since it means there are two oxygen, then put the names together and end the name of the second element with "ide" (remove the last 4 letters).
Silicon + "dI" + ox +"ide"
So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
2NaOH + H₂SO₄ → Na₂SO₄<span>
+ 2H₂O</span>
So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)
If 1000 ml of H₂SO₄ contain 0.750 mol [0.750 M is the amount of moles in
1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol [15 ml is the amount of the acid that took part in the reaction]
⇒
x =
= 0.01125 molMole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
0
2NaOH +
1H₂SO₄ → Na₂SO₄ + 2H₂O
mole ratio of NaOH to H₂SO₄ is 2 : 1∴ if mole of of H₂SO₄ = 0.01125 mol then moles of NaOH = (0.01125 mol) × 2 = 0.0225 molIf 17.5 ml of NaOH contain 0.0225 mol [this was given in the question]
then let 1000 ml of NaOH contain x⇒ x =
= 1.286 mol∴ concentration of NaOH is 1.286 mol/L
V= 50. L n=45 mol T= 200°C = 473k P=?
CP)X 50.L)= (45 mol)(0.0821 light_kimol)(473k)
P = 30am or 4000 kPa