Answer:
125 ml of HCl
Explanation:
The molarity of the stock solution to determine how many milliliters would contain 1.5 moles of HCl. Since a concentration of 12.0 mol/L means that you get 12.0 moles of hydrochloric acid per liter of solution,
Concentration of required HCl (C1) = 1.0M
Volume of required HCL (V1) = 1500 ml
Concentration of stock HCl (C2) = 12M
Volume of stock HCL (V2) = ?
C1V1 = C2V2
V2 = C1V1/C2 = 1*1500/12 = 125 ml
Answer:
The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts using copper salts as reagents or catalysts. It is an example of a radical-nucleophilic aromatic substitution.
Answer:
Xenon
Explanation:
Step 1: Given data
- Pressure (P): 1.00 atm (I looked it up)
Step 2: Convert the temperature to Kelvin
K = °C + 273.15 = 100.0°C + 273.15 = 373.2 K
Step 3: Calculate the number of moles (n)
We will use the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.00 atm × 0.335 L / (0.0821 atm.L/mol.K) × 373.2 K
n = 0.0109 mol
Step 4: Calculate the molar mass of the gas
M = 1.44 g / 0.0109 mol = 132 g/mol
Step 5: Identify the gas
The gas with a molar mass of about 132 g/mol is xenon.
Hydrochloric Acid (HCl) is a strong acid when it is present in the concentrated form but when it is dissolved in water (
) the atoms of this compound dissociates into its respective ions as shown below:

When we add HCl to any complex in its concentrated form the complex does not react at all but when its diluted to 6M and is kept for many hours, the complex reacts slowly. For eg:
![[Co(NH_3)_6]^3^+(aq.) + HCl(aq.) \rightarrow [Co(NH_3)_5Cl]^2^+(aq.) + NH_4^+ (aq.)](https://tex.z-dn.net/?f=%5BCo%28NH_3%29_6%5D%5E3%5E%2B%28aq.%29%20%2B%20HCl%28aq.%29%20%5Crightarrow%20%20%5BCo%28NH_3%29_5Cl%5D%5E2%5E%2B%28aq.%29%20%2B%20NH_4%5E%2B%20%28aq.%29)
As seen from the above reaction it can be seen the positive charge on the complex is reduced by 1 unit because one
ion gets attached to the centre metal atom, therefore we can conclude that the charge on complex gets reduced by 1 unit when HCl reacts with the complex.
Answer:
I will give them advice, that do the power of flame low....
Explanation:
thanks.