M = Sr (strontium)
<em>Step 1.</em> Calculate the <em>moles of CO_2</em>.
Moles of CO_2 = 0.395 g CO_2 × (1 mol CO_2/44.01 g CO_2)
= 0.008 975 mol CO_2
<em>Step 2</em>. Calculate the <em>moles of MCO_3</em>.
Moles of MCO_3 = 0.008 975 mol CO_2 × (1 mol MCO_3/1 mol CO_2)
= 0.008 975 mol MCO_3
<em>Step 3</em>. Calculate the molar mass of <em>MCO_3</em>
MM = grams/moles = 1.324 g/0.008 75 mol = 147.5 g/mol
<em>Step 4</em>. Calculate the <em>atomic mass of M</em>
M_r = <em>x</em> + 12.01 + 3×16.00 = <em>x</em> + 60.01 = 147.5
<em>x</em> = 147.5 – 60.01 = 87.5
<em>Step 5. Identify M</em>.
The element with the closest atomic mass is Sr (A_r = 87.6).
∴ M = Sr and the compound is SrCO_3.
Explanation:
metallic conduction:the movement of electrons through a metal, with no changes in the metal and no movement of the metal atoms.
I hope this help uuhh............
When the total surface area of the solute particles is increased, the solute dissolves more rapidly. Breaking a solute into smaller pieces increases its surface area and increases its rate of solution. So, any answer with “as surface area increases, solid dissolves faster” would be correct. :)
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1. 
2. 
Major Species at C:
1. 
Major Species at D:
1. 
2. 
Major Species at E:
1. 
Major Species at F:
1. 
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK
= 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H
]/[H2
]
pH = pK
= 10.33
Following the key in the diagram (see the attached image), the only particle diagram that represents a mixture of three substances is diagram 2.
To simplify it, let us replace the key in the diagram as follows;
- atom of one element = A
- atom of different element = B
Diagram 1 consists of only AA and AB
Diagram 2 consists of AA, BB, and AB.
Diagram 3 consists of AA and ABA
Diagram 4 consists of AA and BAB
Thus, only diagram 2 has a mixture of 3 substances.
More on mixtures can be found here: brainly.com/question/6594631