Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
Answer:
Molarity = 0.5 M
Osmolarity = 0.5 x 2 = 1 Osmpl.
Molecules of Cl2 = 6.02 x 
/ 4= 1.505 x no. of molecules
Explanation:
If we add half mole in 1L volume than molarity will obviously be 0.5 M.
The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1
Half will be molecules of Ca and half will be of Cl2 for 0.5M.
Answer:
is this based on the newtons law and balnce force
Explanation:
Answer:
The approximate mass of beryllium is 9.0045 u.
Explanation:
Given data:
mass of helium= 4.002 u
mass of beryllium which is 2.25 times of mass of helium= ?
Solution:
we will multiply the mass of helium with 2.25 times,
4.002 u × 2.25 = 9.0045 u
so the mass of beryllium atom would be 9.0045 u.
Answer:
The answer is 0.0698 M
Explanation:
The concentration was prepared by a serial dilution method.
The formula for the preparation I M1V1 = M2V2
M1= the concentration of the stock solution = 0.171 M
V1= volume of the stock solution taken = 200 mL
M2 = the concentration produced
V2 = the volume of the solution produced = 940 mL
Substitute these values in the formula
0.171 × 200 = 490 × M2
34.2 = 490 × M2
Make M2 the subject of the formula
M2 = 34.2/490
M2 = 0.069795
M2 = 0.0698 M ( 3 s.f)
The concentration of the Chemist's working solution to 3 significant figures is 0.0698M
