Answer:
∆H = negative and ∆S = positive.
Explanation:
The reaction given in the question is spontaneous at room temperature ,
hence ,
The the gibbs free energy , i.e. ,∆G will be negative for spontaneous reaction
According to the formula ,
∆G = ∆H -T∆S
The value of ∆G can be negative , if ∆H has a negative value and ∆S has a positive value , because , T∆S , has a negative sign .
Hence , the answer will be , ∆H = negative and ∆S = positive.
Answer:
v = 534.5mL
m = 597.15g
Density = 9.23g/mL
Density = 9.125g/mL
Explanation:
Density = mass/ volume
For the first question
Density = 1.59g/mL
Mass = 834.01g
Volume = ?
Using the above formula we have 1.59 = 834.01/v
v = 834.01/1.59
v = 534.5mL
For the second question
Density =0.9167g/mL
Volume = 651.41mL
Mass =?
Using the above formula we have
0.9167 =m/651.41
Cross multiply
m = 0.9167 x 651.41
m = 597.15g
For the third question
Mass =803.44g
Volume=87.03mL
Density =?
Density = 803.44/87.03
= 9.23g/mL
For the fourth
Density = 56.85/6.23
= 9.125g/mL
Unicellular organisms...
Contain a single cell in their body
Do not contain membrane bound organelles
Use simple diffusion as a transport mechanism
All cellular processes are carried out by the single cell
Only visible under the microscope
Injury in the cell leads to the death of the organism
Asexually reproduce by binary fission
Sexually reproduce by conjugation
Lifespan is comparatively short
Answer:
pH = 2.66
Explanation:
- Acetic Acid + NaOH → Sodium Acetate + H₂O
First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:
- 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
- 1.0 M NaOH * 10.0 mL = 10 mmol NaOH
We<u> calculate how many acetic acid moles remain after the reaction</u>:
- 37.5 mmol - 10 mmol = 27.5 mmol acetic acid
We now <u>calculate the molar concentration of acetic acid after the reaction</u>:
27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M
Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:
- [H⁺] =
Finally we <u>calculate the pH</u>:
