<h3><u> Answer</u>;</h3>
= 4.0 L
<h3><u>Explanation;</u></h3>
Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.
Therefore; <em>Volume α 1/pressure</em>
<em>Mathematically; V α 1/P</em>
<em>V = kP, where k is a constant;</em>
<em>P1V1 = P2V2</em>
<em>V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa</em>
<em>V2 = (0.5 × 203 )/25.4 </em>
<em> = 3.996 </em>
<em> ≈ </em><em><u>4.0 L</u></em>
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
Answer:
Explanation:
mole of HCl remaining after reaction with CaCO₃
= .3 M of NaOH of 32.47 mL
= .3 x .03247 moles
= .009741 moles
Initial HCl taken = .3 x .005 moles = .0015 moles
Moles of HCl reacted with CaCO₃
= .009741 - .0015 = .008241 moles
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O .
1 mole 2 moles
2 moles of HCl reacts with 1 mole of CaCO₃
.008241 moles of HCl reacts with .5 x .008241 moles of CaCO₃
CaCO₃ reacted with HCl = .5 x .008241 = .00412 moles
the mass (in grams) of calcium carbonate in the tablet
= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )
Physical, since you are physically changing its appearance
Answer:
i am new but I haven't reached this level yet
