I think it is 3 because Sus wiw word if rn
Answer:
Explanation:
250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.
150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.
Sodium-25 after 3 minutes : 1.0625 mg
<h3>Further explanation</h3>
General formulas used in decay:

T = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
<h3 />
No=8 mg
t1/2=60 s
T=3 min=180 s

<u>Answer :</u>
Part 13:
The balanced chemical reaction will be:

Part 14:
The balanced chemical reaction will be:

Part 15:
The balanced chemical reaction will be:

<u>Explanation :</u>
Balanced chemical reaction : It is defined as the reaction in which an individual element of an atom present on reactant side must be equal to product side.
Part 13:
The balanced chemical reaction will be:

Part 14:
The balanced chemical reaction will be:

Part 15:
The balanced chemical reaction will be:

Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.