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Setler79 [48]
3 years ago
9

Please help me solve these chemical equations:

Chemistry
2 answers:
Reil [10]3 years ago
8 0
CaCO3 = CaO + CO2
I2 + 2Na2S2O3 = 2NaI + Na2S4O6
6Mg + P4 = 2Mg3P2
2Na + Cl2 = 2NaCl
soldier1979 [14.2K]3 years ago
7 0
This is impossible and confusing!!!
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Consider the formula al2(so4)3. how many atoms of aluminum are present?
Shkiper50 [21]
2 atoms of aluminium
3 0
3 years ago
WILL GIVE BRAINLIEST!!!
Rus_ich [418]

Answer:

The answer is B

Explanation:

The answer is B because representative particles can only be atoms.

3 0
3 years ago
If 2.5L of solution is diluted to prepare 1.7L of a 0.8M solution, what was the original concentration?
mamaluj [8]

Answer:

0.544 M

Explanation:

First find the moles in the final solution

0.8 mols/L *1.7L

1.36 mols

so there is 1.36 mols in 2.5L

concentration will be 1.36/2.5

0.544 M

5 0
2 years ago
Select the correct answer. What energy does a rock resting on a hilltop store? A. gravitational energy B. chemical energy C. the
Colt1911 [192]

Answer:

A.

Explanation:

I chose this because it seems more reasonable . Because its in the air so im guess that its gravitational bc to stay on top of a hill you need balance . Im sorry if its wrong k . I tried .

7 0
3 years ago
Read 2 more answers
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M sodium hydroxide with 0.150 M HBr(aq). (
Taya2010 [7]

Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

  • NaOH → Na+  + OH-

⇒ <em>C </em>NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ <em>C </em>HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ <em>C</em> HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

⇒ pH = 1.21

8 0
3 years ago
Read 2 more answers
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