Answer:
9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.
Explanation:
Molarity : It is defined as the number of moles of solute present in one liter of solution. Mathematically written as:
Moles of citric acid = n
Volume of the citric acid solution = 125 mL =125 × 0.001 L= 0.125 L
(1 mL = 0.001L)
Molarity of the citric acid solution = 0.400 M
n = 0.400 M × 0.125 L = 0.05 moles
Mass of 0.05 moles of citric acid :
9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.
Answer:
16.5 atm
Explanation:
<em>A mixture of He, N₂, and Ar has a pressure of 24.1 atm at 28.0 °C. If the partial pressure of He is 3013 torr and that of Ar is 2737 mm Hg, what is the partial pressure of N₂?</em>
The total pressure of a gaseous mixture is equal to the sum of the partial pressures.
P = pHe + pN₂ + pAr
pN₂ = P - pHe - pAr [1]
We need to express pHe and pAr in atm.
From [1],
pN₂ = 24.1 atm - 3.96 atm - 3.60 atm = 16.5 atm
To determine the mass of gold, we simply multiply the density and volume. Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We do as follows:
mass = density x volume
mass = 19.3 g/cm^3 ( 16.0 cm^3 )
mass = 308.8 g
This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K
V₂ = 9.38 L
You need to list the elements but remember that fluorine has the highest electronegativity out of the entire periodic table
