Answer:
im the only answer your gonna get
Explanation:
Answer: If a substance has a boiling point of 
then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.
Explanation:
The temperature at which vapor pressure of a liquid substance becomes equal to the atmospheric pressure is called boiling point of substance.
At the boiling point, liquid phase and vapor phase remains in equilibrium.
This means that as liquid phase changes into vapor phase and also vapor phase changes into liquid phase at the boiling point.
Thus, we can conclude that if a substance has a boiling point of then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.
Answer:
The pressure remains constant
Explanation:
this is an example in charles law where as the temperature increases so does the volume.
Answer: Volume iron = 2 x 3 x 2 => 12 cm³
Mass = 94 g
D = m / V
D = 94 / 12
D = 7.833 g/cm³
Explanation:
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
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