Answer:
I might be late but when red paint and blue paint mix to become purple. its a <u>physical change</u>, so its C
Explanation:
PKa = -log (Ka) = log [HPO4(2-)] - log[H+]^2 = - log(4.2×10^-13)
pH = - log [H+]
- log [H+]^2 = - 2 log [H+]
2pH = - log (4.2×10^-13) - log [HPO4(2-)]
2pH = - log (4.2×10^-13) - log (0.550)
pH = 6.32
Answer:
T₂ = 259.84 K
T₂ = -13.31 °C
Explanation:
Given data:
Initial pressure = 700 mmHg
Initial temperature = 30.0°C (30+273.15 K = 303.15 K)
Final temperature = ?
Final pressure = 600 mmHg
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
700 mmHg /303.15 K = 600 mmHg / T₂
T₂ = 600 mmHg × 303.15 K / 700 mmHg
T₂ =181890 mmHg.K /700 mmHg
T₂ = 259.84 K
Temperature in celsius
259.84 K - 273.15 = -13.31 °C
Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
