Answer:
Final temperature = 83.1 °C
Explanation:
Given data:
Mass of concrete = 25 g
Specific heat capacity = 0.210 cal/g. °C
Initial temperature = 25°C
Calories gain = 305 cal
Final temperature = ?
Solution:
Q = m. c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C
305 cal = 5.25cal/°C × T2 - 25°C
305 cal / 5.25cal/°C = T2 - 25°C
58.1 °C = T2 - 25°C
T2 = 58.1 °C + 25°C
T2 = 83.1 °C
Answer:
3.861x10⁻⁹ mol Pb⁺²
Explanation:
We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.
We<u> calculate the mass of lead ion in 100 mL of water</u>:
- 100.0 mL ⇒ 100.0 / 1000 = 0.100 L
- 0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²
Now we <u>convert mass of lead to moles</u>, using its molar mass:
- 8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g
- 8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²
Answer:
Potassium selenide
Explanation:
Potassium selenide (K2Se)
