Answer:
Water pressure 0.5 atm
Total Pressure= 2.27 atm
Explanation:
To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.
First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:
P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁
For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:
PV = nRT ∴ P = nRT/V
n will we calculated from the quantity of sample.
2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O
the amount water of hydration is
= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O
= 0.05 mo H₂O
pressure of dry air at the final temperature,
P₂ = 1 atm x 500 K/ 300 K = 1.67 atm
Pressure of water :
P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm
∴ Total Pressure = 1.67 atm
H2O Pressure = 0.5 atm
Answer:
Option D. 5.5
Explanation:
The equation is this:
2A + 6B ⇒ 3C
With the amounts that we were given, let's determine which is the <em>limting reactant</em>
2 A reacts with 6 B
4 A will react with ( 4 .6)/2 = 12B
I have 11 B, so the limiting is B
6 B react with 2 A
11 B will react with (11 .2 )/6 =3.66 A
I have 4 A, so A is the excess.
6 B produce 3 C
11 B will produce ( 11 .3)/6 = 5.5C
Answer:
<em>Di ko</em><em> </em><em>na</em><em> </em><em>po</em><em> </em><em>matandaan</em><em> </em><em>yung</em><em> </em><em>answer</em><em> </em><em>ko</em><em> </em><em>Jan</em><em> </em><em>kasi</em>
<em>nakalimutan</em><em> </em><em>ko</em><em> </em><em>po</em><em> </em><em>sorry</em><em> </em><em>talaga</em>
Explanation:
Answer:
— Molten Potassium Chlorate + sugar (gummi bear) ->
Answer:
0.11M
Explanation:
What is the Molarity of a Ca(OH)2 solution if 30.0mL of
the solution is neutralized by 26.4mL of 0.25M HCI solution?
1L (1000 ml) of the HCl contains 0.25 moles of H ion
26.4 ml contains ( 26.4 X0.25/1000) moles of H ion ion
=0.0066 moles H ion
2HCl +Ca (OH)2-----> 2H2O +CaCl2
SO 2 H IONS NEUTRALIZE 1 Ca(OH)2 MOLECULE
00066 moles H ion neutralize 0.0033 moles Ca(OH)2
the 0.0033 moles are distributed over 30 ml, so the concentration of the
Ca(OH)2 IS 0,0033/(30/1000) =0.11 moles/liter or 0.11M
