Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
Answer:
she can use crystalization method.
Explanation:
She should boil that liquid on flame and then cool it down on mederate temprature and check it out rather the crystals formed or not . if crystals are formed then there will be salts.
And if she want topredict the certain salt then she has to perform certain reactions.
The formula to calculate osmotic pressure is
Osmotic Pressure = M R T
M = Molarity
R = Ideal Gas Constant
T = Temperature in Kelvin
So,
24.6/.2254kg=109.139g /kg >>>>> Molarity
109.139 x mols/92 g = 1.186 mols kg^-1
1.186 x 0.08134 x 298 K = 28.755 atm
<span>1.06852 x 0.08134 x 298K= 26.5 atm
The answer is 26.5</span>
Answer:
pH= 11.49
Explanation:
Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.
From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5
Using the formula below;
[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)
[OH^-] =√(3.2×10^-5 × 0.30M)
[OH^-]= √(9.6×10^-6)
[OH^-]=3.0984×10^-3
pOH= -log[OH^-]
pOH= -log 3.1×10^-3
pOH= 3-log 3.1
pH= 14-pOH
pH= 14-(3-log3.1)
pH= 11+log 3.1
pH= 11+ 0.4914
pH= 11.49
The answer is D . I hope this help you :) .
