Explanation:
The given data is as follows.
Mass of antimony = 19.75 g
Molar mass of Sb = 121.76 g/mol
Therefore, calculate number of moles of Sb as follows.
Moles of Sb = 
= 
= 0.162 mol
Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.
Moles of oxygen = 
= 
= 0.406 mol
Hence, ratio of moles of Sb and O will be as follows
Sb : O
1 : 2.5
We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.
Thus, we can conclude that the empirical formula of the given oxide is
.
Answer: For the elementary reaction
the molecularity of the reaction is 2, and the rate law is rate = ![k[NO_3]^1[CO]^1](https://tex.z-dn.net/?f=k%5BNO_3%5D%5E1%5BCO%5D%5E1)
Explanation:
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction. Thus it can never be fractional.
For elementary reaction
, molecularity is 2 and rate law is ![rate=k[NO_3]^1[CO]^1](https://tex.z-dn.net/?f=rate%3Dk%5BNO_3%5D%5E1%5BCO%5D%5E1)
Water molecules are made of hydrogen atoms and one oxygen atom.
Hope this helped!
Answer:
5.00 g calcium chloride in 95.0 g of water. Solute? ... (4 pts) Calculate the grams of copper (II) nitrate needed to make 100.0 mL of a 3.50 M solution
Explanation:
Both trials of 1.2 g and 1.6 g will have the same mass percent of water because the ratio of the salt to the water of hydration is always constant for any hydrated salt.
<h3>Water of hydration</h3>
For every hydrated salt, the ratio of the salt to the water of hydration remains constant irrespective of the amount of salt taken for experimental analysis.
For example, assuming the mass percent of water in 10g of a hydrated salt is 40%, if 100g of the same salt is taken, the mass percent will remain 40%.
More on water of hydration can be found here: brainly.com/question/11202174