The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J
The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:
7560J= 124g * (100-26)* specific heat
specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
No doubt it's B. solid, liquid, and gas.
Hope this helps!
If the length of a string increases 2 times but the mass of the string remains constant, the original density will be multiplied by a factor mathematically given as
v=1.414 times
<h3>Will the new density of the string equal the original density multiplied by what factor?</h3>
Generally, the equation for the volume is mathematically given as
v = sqt(TL/m)
Where
density = mass/length
Therefore
In conclusion,v ratio will give us
v=1.414 times
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