Answer:
Halogen / salt-former
Explanation:
Bromine is classified as an element in the 'Halogens' section which can be located in group 7 of the Periodic Table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts".
Answer:
Molarity = 1 mol.L⁻¹
Solution:
Molar concentration (Molarity) is given as,
Molarity = Moles / Volume of Solution ------ (1)
Data Given;
Moles = 4 mol
Volume = 4 L
Putting Data in Eq. 1,
Molarity = 4 mol ÷ 4 L
Molarity = 1 mol.L⁻¹
Polar molecules occur when there is an electronegativity difference between the bonded atoms. Nonpolar molecules occur when electrons are shared equal between atoms of a diatomic molecule or when polar bonds in a larger molecule cancel each other out
Answer is true hope I helped!
<span>a) Lead(IV) Carbonate
b) </span><span>327.16
</span><span>
207.2 g + (12.01g + (15.99 x 3) ) x2 = </span><span>327.16 g</span>
The given question is incomplete. The complete question is :
Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
Answer: 28.0 %
Explanation:
To calculate the moles :
![\text{Moles of butane}=\frac{4.65g}{58g/mol}=0.080moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20butane%7D%3D%5Cfrac%7B4.65g%7D%7B58g%2Fmol%7D%3D0.080moles)
![\text{Moles of oxygen}=\frac{10.8g}{32g/mol}=0.34moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20oxygen%7D%3D%5Cfrac%7B10.8g%7D%7B32g%2Fmol%7D%3D0.34moles)
According to stoichiometry :
13 moles of
require 2 moles of butane
Thus 0.34 moles of
will require=
of butane
Thus
is the limiting reagent as it limits the formation of product and butane is the excess reagent.
As 13 moles of
give = 10 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Thus 0.34 moles of
give =
of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Mass of ![H_2O=moles\times {\text {Molar mass}}=0.26moles\times 18g/mol=4.68g](https://tex.z-dn.net/?f=H_2O%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.26moles%5Ctimes%2018g%2Fmol%3D4.68g)
![{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bpercentage%20yield%7D%7D%3D%5Cfrac%7B%5Ctext%20%7BExperimental%20yield%7D%7D%7B%5Ctext%20%7BTheoretical%20yield%7D%7D%5Ctimes%20100%5C%25)
![{\text {percentage yield}}=\frac{1.31g}{4.68g}\times 100\%=28.0\%](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bpercentage%20yield%7D%7D%3D%5Cfrac%7B1.31g%7D%7B4.68g%7D%5Ctimes%20100%5C%25%3D28.0%5C%25)
The percent yield of water is 28.0 %