It has a high specific heat. Meaning it takes a lot of heat for it to raise up 1 degree
Answer:
52.9 KJmol-1
Explanation:
From;
log(k2/k1) = Ea/2.303 * R (1/T1 - 1/T2)
The temperatures must be converted to Kelvin;
T1 = 25° C + 273 = 298 K
T2= 35°C + 273 = 308 K
R= gas constant = 8.314 JK-1mol-1
Substituting values;
log 2 = Ea/2.303 * 8.314 (1/298 - 1/308)
Ea = 52.9 KJmol-1
Answer: B
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Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get

To calculate the volume of acid, we use the equation given by neutralisation reaction:

where,
are the molarity and volume of stock acid which is 
are the molarity and volume of dilute acid which is 
We are given:

Putting values in above equation, we get:

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid