Answer:
87.3 calories of heat is required.
Explanation:
Heat = mcΔT
m= mass, c = specific heat of silver, T = temperature
H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)
H = 57.8 * 0.057 * 26.5
H = 87.3069 cal.
The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.
Answer:
+3
Explanation:
The oxygen all have a -2 oxidation state. (peroxides are exceptions)
The chemical structure is symmetrical. Both carbon are equivalent.
2 (oxidation state of carbon) + 4 (oxidation state of oxygen) = charge of ion.
2 (oxidation state of carbon) + 4 (-2) = -2
oxidation state of carbon = +3
Answer:
The total amount of heat released is 68.7 kJ
Explanation:
Given that:
mass of water = 94.0 g
moles of water = 94 / 18.02 = 5.216
80⁰C ------> 0⁰C --------> -30⁰C
Q1 = m Cp dT
= 94 x 4.184 x (0 - 80)
= -31463.68 J
= -31.43 kJ
Q2 = 6.01 x 10^3 x 5.216
= - 31348.16 J
= -31.35 kJ
Q3 = - 94 x 2.09 x 30
= - 5893.8 J
= -5.894 kJ
Total heat = Q1 + Q2 + Q3 = -31.43 kJ + (-31.35 kJ ) + (-5.894 kJ
) = -68.7 kJ
Total heat released = -68.7 kJ
Note that the "negative sign" simply indicates heat released, therefore no need to put it in the answer.
Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T = 
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where, 
= concentration =
Hence, putting the given values into the above equation as follows.
= 
= 
= 
m
or, = 
= 1 nm (approx)
Also, it is known that = 
Hence, we can conclude that addition of 0.1 of KCl in 0.1 of NaBr "" will decrease but not significantly.
