Specific heat capacity of any substance comes with the unit : J/(g*degree C)
for molar capacity , change gram -> moles unit ( J / moles * degree C)
4.18 J / mol - degree C
H = 1.01 g * 2 = 2.02 g
O = 16 g
2.02 + 16 = 18.02 g
Now :- 4.18 J / mol- degree C) * 18.02 / 1 mole H2O
molar heat = 75.3 J / mol - degree C
<span />
Answer:
59.92 × 10²³ atoms are in 9.95 moles of iron
1.8 ×10²² molecules are in 0.03 moles of Carbon dioxide
1.19 moles are found in 7.20 x 10^23 atoms of platinum
The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
Answer:
1. Ba2+ 2. Sr2+
Explanation:
When a solution contains the Barium ,Ba²⁺ ion or Strontium, Sr²⁺ ion, they reacts with either H₂SO₄(aq) or Na₂SO₄(aq) to produce a white precipitate of BaSO₄(s) and SrSO₄(s) respectively
The chemical reactions are given below
Ba²⁺ + H₂SO₄(aq) ⇒ BaSO₄(s) + 2H⁺ (aq)
Ba²⁺ + Na₂SO₄(aq) ⇒ BaSO₄(s) + 2Na⁺ (aq)
Sr²⁺ + H₂SO₄(aq) ⇒ SrSO₄(s) + 2H⁺ (aq)
Sr²⁺ + Na₂SO₄(aq) ⇒ SrSO₄(s) + 2Na⁺ (aq)
