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Gnom [1K]
3 years ago
15

Science is a continually ongoing process with many cycles. True False

Chemistry
1 answer:
Lina20 [59]3 years ago
6 0

Answer true

Explanation:becoz it's true

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I need help on homework
My name is Ann [436]
What do you need help with
3 0
3 years ago
Classify each of the following reactions as either exothermic or endothermic: 2H,O(l) + heat → 2H₂(g) + O₂(g)​
Murljashka [212]

Answer:

endothermic

Explanation:

It is endothermic since it absorbs heat.  If the reaction RELEASED heat (exothermic) the "+ heat" would be on the right side of the equation.

6 0
1 year ago
2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g
julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system.  Now;

Number of moles of Fe =  12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with  0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of  Fe3O4(s)

 0.22 moles  of O2   produces   0.22 moles * 1 mole/3 moles of  Fe3O4(s) = 0.1073 moles

Mass of  Fe3O4(s) =0.1073 moles  * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0
2 years ago
4. What type of evidence do you think would be most difficult to collect
mars1129 [50]

I think it is trace evidence since it is really small and hard to find.

3 0
3 years ago
Read 2 more answers
Will a double replacement reaction occur if a potassium hydroxide solution is mixed with a lead(II) nitrate solution? Briefly ex
ohaa [14]
No - a precipitation will occur though. Potassium nitrate is soluble in water, so the potassium and nitrate ions will remain spectator ions and stay in solution. Lead (II) hydroxide is not soluble, and will precipitate out of solution to form a solid product.
3 0
3 years ago
Read 2 more answers
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