Density = Mass divided by Volume
So what you do is 80.00 ÷ 0.99755.
Which I believe would equal <span>80.1964813794
Hope this helps! </span>
According to Dalton's Law, in a mixture of non-reacting gasses, thetotal pressure<span> exerted is the sum of the </span>partial pressures<span> of the component gasses. In more complicated circumstances, equilibrium states come into effect, but fortunately for us, </span>oxygen<span> is non-reactive with </span>water vapor<span>.</span>
Answer:
The answer to your question is: 16.7 g of KBr
Explanation:
Data
mass KBr = ? g
Volume = 0.400 L
Concentration = 0.350 M
Formula
Molarity = moles / volume
moles = molarity x volume
Process
moles = (0.350)(0.400)
= 0.14
MW KBr = 39 + 80 = 119 g
119 g of KBr -------------------- 1 mol
x -------------------- 0.14 mol
x = (0.14 x 119) / 1
x = 16.7 g of KBr
Because pure silicon is a perfect semiconductor.
For room temperature, it rarely conducts, you can search for the threshold temperature, the characteristic equation is fairly complicated.
Answer:
When halogen elements react with group one metals they form halide salts.
Explanation:
The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while iodine is solid.
When halogen elements react with group one metals they form halide salts.
Alkali metals have one valance electron and halogens needed one electron to complete the octet thus alkali metals loses one electron which is accepted by halogens atom and form ionic compound called halide salts.
For example:
2Na + Cl₂ → 2NaCl
2K + Cl₂ → 2KCl
2Rb + Cl₂ → 2RbCl
2Li + Cl₂ → 2LiCl
With bromine:
2Na + Br₂ → 2NaBr
2K + Br₂ → 2KBr
2Rb + Br₂ → 2RbBr
2Li + Br₂ → 2LiBr
With iodine:
2Na + I₂ → 2NaI
2K + I₂ → 2KBI
2Rb + I₂ → 2RbI
2Li + I₂ → 2LiI
