Answer:
( About ) 5.7 grams
Explanation:
Take a look at the attachment below for a proper explanation;
Answer:
The answer to your question is T2 = 610.9 °K
Explanation:
Data
Volume 1 = V1 = 30 cm³
Temperature 1 = T1 = 420 °K
Pressure 1 = P1 = 110 kPa
Volume 2 = V2 = 40 cm³
Temperature 2 = T2 = ?
Pressure 2 = P2 = 120 kPa
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for T2
T2 = P2V2T1 / P1V1
-Substitution
T2 = (120 x 40 x 420) / (110 x 30)
-Simplification
T2 = 2016000 / 3300
-Result
T2 = 610.9 °K
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
Learn more about pH of buffer:
brainly.com/question/21881762
Hci acid its high preety cool
Answer:
Explanation:
Mass, m = 1 kg
The density of water, d = 1000 kg/m³
We need to find the volume of water. The formula for the density of an object is equal to mass divided by volume. Its formula is :
d = m/V
V is the volume of water
So, the volume of water is 
.
