The answer is A) Aluminum chloride - It does not have the prefix "tri-" since it is ionic.
The balanced chemical equation of the reactions given is as follows:
- 2LiHCO3 -----> Li2CO3 + H2O + CO2
- 2 N2 + 5 O2 -----> 2 N2O5
- MgBr2 + KOH -----> KBr + Mg(OH)2
- Mn + 2 CuCl -----> 2 Cu + MnCl2
- 8 Zn + S8 -----> 8 ZnS
- 2 NaOH + H2SO4 -----> 2 H2O + Na2SO4
- 2 K + 2 H2O -----> 2 KOH + H2
- C5H12 + 8 O2 -----> 6H2O + 5 CO2
- 2 KOH + H2CO3 -----> 2 H2O + K2CO3
- C4H802 + 6 O2 -----> 4 H20 + 4 CO2
- 16 Al + 3 S8 ---> 8 Al2S3
<h3>How to balance chemical equations</h3>
Balancing of chemical equations is the process of adding numerical coefficients in front of moles of reactants and products to ensure that the moles of atoms of elements of the reactants is equal to the moles of atoms of products formed.
The balanced chemical equation of the reactions given is as follows:
- 2LiHCO3 -----> Li2CO3 + H2O + CO2
- 2 N2 + 5 O2 -----> 2 N2O5
- MgBr2 + KOH -----> KBr + Mg(OH)2
- Mn + 2 CuCl -----> 2 Cu + MnCl2
- 8 Zn + S8 -----> 8 ZnS
- 2 NaOH + H2SO4 -----> 2 H2O + Na2SO4
- 2 K + 2 H2O -----> 2 KOH + H2
- C5H12 + 8 O2 -----> 6H2O + 5 CO2
- 2 KOH + H2CO3 -----> 2 H2O + K2CO3
- C4H802 + 6 O2 -----> 4 H20 + 4 CO2
- 16 Al + 3 S8 ---> 8 Al2S3
Learn more about balancing of chemical equations at: brainly.com/question/15428811
Freeze-Thaw Weathering
This action can widen the cracks in the rock, and when the temperature rises above freezing, the ice thaws, allowing the water to seep further into the cracks. As this process of freezing and thawing happens repeatedly, the rock begins to weaken and eventually breaks apart into angular fragments.
Answer:
Explanation:
10 mL = .01 L .
25 mL = .025 mL .
10 mL of .1 M NaOH will contain .01 x .1 = .001 moles
25 mL of .1M HCl will contain .025 x .1 = .0025 moles
acid will neutralise and after neutralisation moles of acid remaining
= .0025 - .001 = .0015 moles .
Total volume = .01 + .025 = .035 L
concentration of remaining HCl = .0015 / .035
Option D is correct.
= .042857 M
= 42.857 x 10⁻³ M .
pH = - log [42.857 x 10⁻³]
= 3 - log 42.857
= 3 - 1.632
= 1.368 .
