a sample of solid is decomposed and found to contain 6.52g of potassium, 4.34 g of chromium and 5.34 of oxygen, what is the empi
1 answer:
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Answer: 502 Joules
Explanation:
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 40.0 mL
Putting values in above equation, we get:
When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
q = heat absorbed by water
= mass of water = 40.0 g
= final temperature of water = 20.0°C
= initial temperature of water = 17.0°C
= specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
Hence, the joules of heat were re-leased by the lead is 502
Answer:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u> The final concentration of sodium anion = 0.1035 M</u>