Taking into account the scientific notation, the result of the sum is 10.84300×10³.
- <u><em>Scientific notation</em></u>
First, remember that scientific notation is a quick way to represent a number using powers of base ten.
The numbers are written as a product:
a×10ⁿ
where:
- a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
- n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.
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<u><em>Sum in scientific notation</em></u>
You want to add two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 3.
Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 9.7300×10²= 0.97300×10³
Taking the quantities to the same exponent, all you have to do is add what was previously called the number "a". In this case:
0.97300×10³ + 9.8700×10³= (0.97300+ 9.8700)×10³= 10.84300×10³
Finally, the result of the sum is 10.84300×10³.
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Answer: the reliability will be worse
Explanation:
Suppose we used 0.5 M NaOH to titrate our vinegar sample instead of 0.1 M.
Now by using 0.5M instead of 0.1M we are increasing the concentration of NaOH,
We know that the moles used = Volume x concnetration.
so for the same no of moles, if the concentration increases, the volume decreases.
Hence it will consume less NaOH.
now Since the volume decreases, the titration volume of less number will increase the % error.
Therefore the reliability will be worse.
The position of equilibrium lies far to the right, with products being favored.
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
