Answer:
869.93J/g°C
Explanation:
From the question given, we obtained the following:
M = 42.6g
T1 = 12°C
T2 = 46°C
ΔT = T2 — T1 = 46 — 12 = 34°C
C =?
Q = 1260 KJ = 1260000J
Q = MCΔT
C = Q / MΔT
C = 1260000 /(42.6 x 34)
C = 869.93J/g°C
Therefore, the specific heat capacity is 869.93J/g°C
Answer: Heterogeneous type of mixture.
Explanation:
This type of mixture are easily be seen and one may need a microscope sometimes. This category includes the suspensions where visible particles settles.
Answer:
Mass of iron = 59.375 gm
Explanation:
Calories ( or joules) are added to the water by the hot steel so at the endpoint they are BOTH at 40 C
The water gains:
4.18 j/g-C * 50 * (40-30 C) = 2090 j
The steel gave up 2090 j going from 120 to 40 C
2090 = .44 j/g-C * m * (120-40) solve fro m = 59.375 gm
the percent yield of the reaction is 100%.
The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:
% yield = actual yield / theoretical yield * 100%
% yield of a reaction in this case Rate
In this case, the molar mass of NaBr is 102.9 g / mol, as you know:
444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g
theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g
, Replaced by the definition of percent yield:
percent yield = 728.532 grams / 728.532 grams * 100%
percent yield = 100%
Finally, the percent yield of the reaction is 100%.
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FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.
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