C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol
mKI: 39+127 = 166 g/mol
1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI
:)
Answer:
<u>89.6 L</u>
Explanation:
In normal conditions,
<u><em>For every </em></u><u><em>1 mole</em></u><u><em> of carbon dioxide at STP, it occupies </em></u><u><em>22.4 L</em></u><u><em> of volume.</em></u>
<u><em /></u>
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Solving :
⇒ 1 mole : 22.4 L
⇒ 1 × 4 : 22.4 × 4
⇒ 4 moles : <u>89.6 L</u>
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The heat released by reaction : C) -8870 J
<h3>Further explanation</h3>
Given
1.008 g of hydrogen
500.00 g water
The temperature rises 25.00 °C to 29.24 °C
Required
energy required
Solution
Q absorbed by water :
Q = m.c.Δt
Q = 500 g x 4.18 J/g C x (29.24-25)
Q = 8870.08 J
The reaction to produce HCl is an exothermic reaction (releasing heat), so that Q is negative
Q water = -Q HCl = -8870.08 J
Answer:
The solid will sublime into a gas
Explanation:
In the graph in the attachment
Y-axis represents pressure in mm of Hg
X- axis represents temperature in degree Celsius
The graph show the triple point of water.
Clearly, the sample will go from solid to vapor directly, no liquid involved (pink line is there to show that this is below triple point)
So
expect sublimation
The solid will sublime into a gas.
