1) You need to use the atomic mass of copper.
You can find it in a periodic table. It is 63.546 amu.
2) The atomic mass is the weigthed mass of the different isotopes.
This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:
=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.
3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x
=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278
=> 63.546 = 62.9296x + 64.9278 - 64.9278x
=> 64.9278x - 62.9296 = 64.9278 - 63.546
=> 1.9982x = 1.3818
=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%
=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%
Answer:
Cu-63 69.15%;
Cu-65 : 30.85%
The volume of the liquid in this diagram shown above would be equal to 36.5 mL.
<h3>What is a
graduated cylinder?</h3>
A graduated cylinder is also known as measuring cylinder and it can be defined as a narrow, cylindrical piece of laboratory equipment with marked lines, which are used to measure the volume of a liquid.
In order to take a reading for the measurement of the volume of a liquid such as water, you should ensure that your eye level is even with the center of the meniscus.
In this scenario, the volume of the liquid in this diagram would be 36.5 mL because each of the small lines on the graduated cylinder measures 0.5 mL.
Read more on graduated cylinder here: brainly.com/question/24869562
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We are given –
- Final velocity of car is, v= 0
- Initial velocity of car is, u= 100 km/hr
- Time taken, t is = 3 minutes or 180 sec
Here–
Now –
____________________________
_______________________________
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
