Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Answer:
D.) They often form hydroxide ions.
Explanation:
They generate hydroxide ions in water. they are soapy to touch and bitter in taste. they conduct electricity.
(all bases have hydroxide ions)
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
A Beta particles is emitted when an atom of 85Kr spontaneously decays.
Answer:
C. 1 proton 3 neutrons
Explanation:
A nucleus is more stable if the ratio of the neutrons to protons is between 1:1 and 1:1.5.
Thus the ratios of neutrons to protons for the nuclei are as follows
A- 1:1
B- 1:1
C- 1:3
D- 1:2
Among these ratios, C is the greatest thus the nucleus is the least stable.
