To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, the equation becomes 186.207=187*0.626+185*x where x is the percent abundance of 185. The answer is 0.374 or 37.4%. This can also be obtained by 100%-62.6%= 37.4%.
Answer:
No, i will not use a water pipe consisting of the two metals
Explanation:
Looking at the reduction potential of the both metals, it is clear that an electrochemical cell is set up with iron as the anode and copper as the cathode.
This will make the iron to quickly corrode and eventually destroy the water pipe. It is better to have a set up in which another metal that is higher than iron in the electrochemical series is combined with it.
<h3>
Answer:</h3>
0.34 mol S
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
11 g S
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of S - 32.07 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
0.343 mol S ≈ 0.34 mol S
I couldn't really find anything about the growth time but it does say that it could remain viable in soil for up to 40 years
Answer:
40.02 calories
Explanation:
V = 10 mL = 10g
we know t went <em>up</em> by 4°C, this is our ∆t as it is a change.
Formula that ties it together: Q = mc∆t
where,
Q = energy absorbed by water
m = mass of water
c = specific heat of water (constant)
∆t = temperature change
Q = (10 g) x (4.186 J/g•°C) x (4°C)
Q = 167.44 J
Joules to Calories:
167.44 J x 1 cal/4.184 J = 40.02 calories
(makes sense as in image it is close to the value).