Answer:
-209 kJ
Explanation:
I did the math. You're welcome ;)
To solve this question, you must use the formula: q=mc(change in temperature), where q is heat, m is mass, C is specific heat and temperature change is temperature change. The specific heat for ice is 2.1kJ/Kg x K (given). The change in temperature is 15 degrees Celsius (which you should change to kelvins so you can cancel out units), or 273 + 15 = 288K. The mass is 150 grams, which is 0.15 kg. Now, we can solve for q, heat. We will do this by substituting variables into the formula. After simplifying and cancelling out units, the answer we get is: 90.72kJ.
Answer:
A. H3O+/OH−
Explanation:
A conjugate acid-base pair are a pair of molecules that differ in 1 H⁺
A. The conjugate pair of H₃O⁺ = H₂O not OH⁻
B. The conjugate pair of NH₄⁺ is NH₃
C. The conjugate pair of C₂H₃O₂⁻ is HC₂H₃O₂
D. The conjugate pair of H₂SO₃ is HSO₃⁻
That means right option, that is not a conjugate acid-base pair, is:
<h3>A. H3O+/OH−
</h3>
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
