Answer:
The specific heat of the alloy 
Explanation:
Mass of an alloy 
= 25 gm
Initial temperature 
= 100°c = 373 K
Mass of water 
= 90 gm
Initial temperature of water 
= 25.32 °c = 298.32 K
Final temperature 
= 27.18 °c = 300.18 K
From energy balance equation
Heat lost by alloy = Heat gain by water
[ - ] = 
( - )
25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)
This is the specific heat of the alloy.
Answer:
11,000 cm
Explanation:
Step 1: Given data
Width of the field (w): 17 meters
Length of the field (l): 38 meters
Step 2: Calculate the perimeter of the field
The field is a rectangle. We can find its perimeter (P) by adding its sides.
P = 2 × w + 2 × l = 2 × 17 m + 2 × 38 m = 110 m
Step 3: Convert the perimeter to centimeters
We will use the relationship 1 m = 100 cm.
110 m × (100 cm/1 m) = 11,000 cm
Answer:
See explanation below
Explanation:
In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.
At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.
Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.
In the picture below, you have the mechanism for all the 4 products.
Hope this helps
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
Answer:
Molar heat of solution of KBr is 20.0kJ/mol
Explanation:
Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.
The dissolution of KBr is:
KBr → K⁺ + Br⁻
In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:
q = 1.36kJK⁻¹ × 0.370K
q = 0.5032kJ
Moles of KBr in 3.00g are:
3.00g × (1mol / 119g) = 0.0252moles
Thus, molar heat of solution of KBr is:
0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>
