Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:
1. Weigh all the equipment and materials required in the experiment before the experiment.
2. Avoid spillage and evaporation during the experiment.
3. Weigh all the equipment and materials after the experiment.
If the mass is conserved then weight from step 1 is equal to weight from step 3.
Answer:
An acid is a substance that donates protons (in the Brønsted-Lowry definition) or accepts a pair of valence electrons to form a bond (in the Lewis definition). A base is a substance that can accept protons or donate a pair of valence electrons to form a bond.
Hope it helps!!!
<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M
<u>Explanation:</u>
We are given:
Initial concentration of chlorine gas = 0.0300 M
Initial concentration of bromine monochlorine = 0.0200 M
For the given chemical equation:
<u>Initial:</u> 0.02 0.03
<u>At eqllm:</u> 0.02-2x x 0.03+x
The expression of 
for above equation follows:
![K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBr_2%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BBrCl%5D%5E2%7D)
We are given:
Putting values in above equation, we get:
Neglecting the value of x = -0.96 because, concentration cannot be negative
So, equilibrium concentration of bromine gas = x = 0.00135 M
Hence, the equilibrium concentration of bromine gas is 0.00135 M
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
