Formula : BaI₂. <span>barium iodide</span>
Answer: AgCl + Fe
Explanation:
A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.
A general single displacement reaction can be represented as :

The blanced chemical reaction will be :

What's the problem ? Hardness is not the definition of a metal.
You need to expand your thinking. EVERY element is solid, liquid, and gas, over different ranges of temperature ... including all of the metals. There are only TWO elements that are liquid AT ROOM TEMPERATURE, and mercury is one of them. But on a mild day at the south pole, mercury is solid too.
Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:

where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
![\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B515%7D%7B760%7D%3D%5Cfrac%7B%5CDelta%20H%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B429K%7D-%5Cfrac%7B1%7D%7B415K%7D%5D)

Thus the heat of vaporization is 41094 Joules
Answer:
1.26 M
Explanation:
The ion nitrate is NO₃⁻ and the Barium is from group 2 so it forms the ion Ba²⁺, so the barium nitrate has the formula: Ba(NO₃)₂. The molar masses are: Ba: 137 g/mol, N = 14 g/mol, O = 16 g/mol, so the molar mass of barium nitrate is:
137 + 2x(14 + 3x16) = 199 g/mol
The number of moles is the mass divided by the molar mass, so:
n = 25.1/199 = 0.126 mol of Ba(NO₃)₂
In 1 mol of the salt, there are 2 moles of NO₃⁻, so the number of moles of nitrate is 0.252 mol. Nitrates formed with ammonium (that can react when the solid dissolves) and with elements from group 1 and 2 are completely soluble in water. So, the moles of nitrate will remain 0.252 mol.
The molarity is the number of moles divided by the volume (0.2 L):
[NO₃⁻]= 0.252/0.2 = 1.26 M