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Tomtit [17]
2 years ago
6

Help on this question please

Chemistry
2 answers:
dedylja [7]2 years ago
5 0

Answer:

Choice 1: Position 3

Explanation:

The equinox is showed in Position 3 and 1.

Position 1 is fall, Position 2 is Winter, 3 is spring, 4 is summer. Since you are looking for the northern hemisphere, the shadowed area shown is position 2 is winter, spring is next in rotation so that already gives away the fact that position 3 is spring.

garri49 [273]2 years ago
4 0
Position 3 is the answer.
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To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7
Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0
2 years ago
Dissolving continues until an equilibrium is established.<br> O True<br> O False
vazorg [7]
Answer: It’s true

Explanation:
7 0
3 years ago
What mass of barium carbonate can be produced when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a
valentinak56 [21]
100M solution so did new wells alanna annan carbonated
3 0
2 years ago
If I have 3.5 moles of C, and excess Fe2O3 , how many moles of Fe can I produce?
Wittaler [7]

Answer:

3.5 moles Fe

Explanation:

From the equation, Reaction of 2 moles of Fe₂O₃ with  1 mole of C produces 1 mole of Fe. When excess Fe₂O₃ is used, the only liming factor is C.

The ratio of amount of C used to the amount of Fe produced is 1:1

Therefore, if 3.5 moles of C are used,  3.5 moles of Fe are also produced.

6 0
3 years ago
Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)
Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation  Cr(s) → Cr3+ + 3e-  

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the  half-reactions

2(Cr → Cr3+ + 3e-)    E° ox = 0.74 V

3(Ni2+ +2e- → Ni)     E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox  = -0.25 + 0.74  = 0.49

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

b)

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution)   MnO4-  +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e-   (oxidation)   Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the  half-reactions

MnO4-  +8H+ +10-e- ⇔ Mn2+ +4H2O    E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e-        E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox  = 1.51 + 2.12  = 3.63 V

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0
3 years ago
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