Answer:
390
Explanation:
Specific heat capacity= heat/mass × temperature

Remember you convert gram into kilogram and 1 gram =0.001 kilogram
Answer:
C. Arrhenius
An Arrhenius base is a substance that dissociates in water to form hydroxide (OH–) ions. In other words, a base increases the concentration of OH– ions in an aqueous solution.
Explanation:
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Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.