If it is cooled the motion of the particles decreases as they lose energy.
Multiply the mass by the abundance and add each. Make sure to convert the percentage into a decimal. (49.946 * .043) + (51.941 * .838) + (52.941 * .095) + (53.939 * .024) = 51.99 round up using sig figs and the answer is c. 52.00 amu
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
Answer:
27 liters of hydrogen gas will be formed
Explanation:
Step 1: Data given
Number of moles C = 1.03 moles
Pressure H2 = 1.0 atm
Temperature = 319 K
Step 2: The balanced equation
C +H20 → CO + H2
Step 3: Calculate moles H2
For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2
For 1.03 moles C we'll have 1.03 moles H2
Step 4: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of the H2 gas = 1.0 atm
⇒with V = the volume of H2 gas = TO BE DETERMINED
⇒with n = the number of moles H2 gas = 1.03 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 319 K
V = (n*R*T)/p
V = (1.03 * 0.08206 *319) / 1
V = 27 L
27 liters of hydrogen gas will be formed
