The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL
Now, we use Boyle's law, that for a given amount of gas,
PV = constant
P x 6843 = 101.3 x 83776
P = 1,240 kPa
Mass of water added:
0.997 x 1500
= 1495.5 grams
a) Volume = mass / density
Volume = 1495.5 / 0.917
Volume = 1630 cm³ = 1.63 L
b) The ice cannot be contained in the bottle as its volume exceeds that of the bottle.
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
A polar molecule<span> has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from </span>polar<span> bonds arranged asymmetrically. Water (H</span>2<span>O) is an example of a </span>polar molecule<span> since it has a slight positive charge on one side and a slight negative charge on the other.</span>
Explanation:
The given data is as follows.
= 1.5 atm, 
= 20 L, 
= (28 + 273) K = 301 K
= 5 atm, 
= ?, 
= (50 + 273) K = 323 K
Formula to calculate the volume will be as follows.
= 
Putting the given values into the above formula as follows.
=
= 
= 0.64 L
Thus, we can conclude that the change in volume of the balloon will be 0.64 L.
