When 67 g of water is heated from its melting point to its boiling point, it takes 28006 J of heat.
<h2>Relationship between heat production and temperature change</h2>
- A way to numerically relate the quantity of thermal energy acquired (or lost) by a sample of any substance to that sample's mass and the temperature change that results from that is provided by specific heat capacity.
The following formula is frequently used to describe the connection between these four values.
q = msΔT
where, q = the amount of heat emitted or absorbed by the thing
m = the object's mass = 67 gm
s = a specific heat capacity of the substance = 4.18 J/gC
ΔT = the resultant change in the object's temperature = 373.15 -273.15K= 100 k
q = 67 * 4.18 * 100 J
⇒q = 28006 J
Therefore it is concluded that 67 g of water takes 28006 J of heat from its melting point to reach its boiling point.
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Answer:
No
Explanation:
I'm not educated enough on the matter but from what I've been taught water boils at 100 Celsius and it simultaneously evaporates.
Answer:
Explanation:
Noble gases occupy the last group of the periodic table
They have fully filled valence electron shell
What this means is that they have attained stability and thus do not take part in chemical bonding that usually invloves the transfer or sharing of electrons
Thus, noble gases are snobs because they do not partake in chemical bonding with atoms of other elements or atoms of theirselves
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
<em>Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.</em>
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
<em>Moles CH3COOH: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
<em>Moles CH3COONa: </em>
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
<h3>pH = 4.71</h3>
