Explanation:
The observation of student was that thermometer reading changed from 27°C to 35°C which indicates that temperature of the beaker solution rose after reaction due to release of heat during reaction as a product.
Those chemical reactions which gives heat energy as a product into their surrounding are categorized as exothermic reactions. During the course of these reaction temperature of the surroundings also increased.
So, this means that reaction between silver nitrate and copper wire is an exothermic reaction.
Three ways that the student could speed up the reaction :
- By adding catalyst to the reaction.
- By decreasing the temperature.
- By increasing the concentration of silver nitarte solution.
Answer:
Salts and acids are examples of inorganic compounds called <u><em>electrolytes</em></u>.
Explanation:
Electrolytes are the substances which dissociates into ions when dissolved in water and due to this they are able to conduct electric current through them. These compounds in solid form does not conduct electricity due to the absence of free ions.
For example: Sodium chloride , sulfuric acid etc.


Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>