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:
4d orbital.
Explanation:
Hello!
In this case, since zirconium's atomic number is 40, we fill in the electron configuration up to 40 as shown below:
Thus, the orbital 4d is partially filled.
Best regards!
<u>Answer:</u> The average atomic mass of X is 28.09 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Mass of isotope 1 = 27.979 amu
Percentage abundance of isotope 1 = 92.21 %
Fractional abundance of isotope 1 = 0.9212
Mass of isotope 2 = 28.976 amu
Percentage abundance of isotope 2 = 4.70 %
Fractional abundance of isotope 2 = 0.0470
Mass of isotope 3 = 29.974 amu
Percentage abundance of isotope 3 = 3.09 %
Fractional abundance of isotope 3 = 0.0309
Putting values in equation 1, we get:
![\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20X%7D%3D%5B%2827.979%5Ctimes%200.9212%29%2B%2828.976%5Ctimes%200.0470%29%2B%2829.974%5Ctimes%200.0309%29%5D)
Hence, the average atomic mass of X is 28.09 amu
Answer:
Your answer should be 15.68 grams.
Explanation:
Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.
