Explanation:
Solids have definite shape and distinct boundaries as such the intermolecular forces between them is maximum
• Before the balloon was placed inside the hot water, the pressure was the same inside and outside the balloon. The hot water raised the kinetic energy of the air molecules inside the balloon, expanding the balloon, through thermal expansion.
• (1) the pressure of air inside the balloon increased, (2) the volume of the inside of the balloon increased as well, and (3) the temperature of the balloon increased. Note that pressure and volume are inversely proportional, and pressure and temperature are directly proportional. Therefore as the temperature increases, the pressure inside will increase, causing an increase in the volume. At a certain point though the volume will increase too much as to cause a significant decrease in pressure.
• The air molecules will gain kinetic energy, hence (1) increasing the molecules's speed, and (2) heating the air molecules.
increases my factor of 10
Answer:
0.289J of heat are added
Explanation:
We can relate the change in heat of a substance with its increasing in temperature using the equation:
q = m*ΔT*S
<em>Where Q is change in heat</em>
<em>m is mass of substance (In this case, 0.0948g of water)</em>
<em>ΔT = 0.728°C</em>
<em>S is specific heat (For water, 4.184J/g°C)</em>
Replacing:
q = 0.0948g*0.728°C*4.184J/g°C
q = 0.289J of heat are added
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
