Answer : If a substance is the limiting reactant, then it limits the formation of products because in the reaction it is present in limited amount.
Explanation :
While observing a chemical reaction, we can tell about whether a reactant is limiting or excess.
Step 1 : first write the chemical reaction and then balanced the chemical equation.
Step 2 : convert the given masses into the moles if mass of 
is 10.5 g and molar mass of is 28 g/mole and the mass of hydrogen is 0.40 g and molar mass of hydrogen is 2 g/mole.
Step 3 : Now we have to determine the limiting reagent and excess reagent.
Now we conclude that 
is the limiting reagent and hydrogen is an excess reagent.
Hypothesis :
Limiting reagent : It is the reagent in the chemical reaction that is totally consumed when the chemical reaction is complete. Limiting reagent limits the formation of products.
The moles of magnesium in 1.25 x 1023 are 20.8 mol.
Answer:
The three statements are true
Explanation:
For the reaction:
I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)
State oxidation of iodine in I₂O₅ is:
5 O²⁻ = 10⁻
As you have 2 I and the molecule has no charge, <em>oxidation state of I is +5</em>.
The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means <em>the carbon is oxidized</em>
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An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being <em>the oxidizing agent</em>
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Thus,<em> the three statements are true</em>.
<h2>Answer:</h2>
C) The scientist made an identification
<h2>Explanation:</h2>
The fingerprint is a very certain method for identifying a person, because all human beings have got unique fingerprints. Forensic laboratories have database of all citizens in the area. Hence an identification is done by the results matched with the database information. The two main categories of fingerprint matching techniques are minutiae-based matching and pattern matching and both are them are present in the given database.
Answer:
58mL
Explanation:
Given parameters:
Density of water = 1g/mL
Mass of object = 58g
Unknown:
The volume the object must have to be able to float in water = ?
Solution:
To solve this problem, we know that the object must have density value equal to that of water or less than that of water to be able to float.
We then set its density to that of water;
Density = 
Volume = 
So;
Volume = 
= 58mL
