The rounding up of the aforementioned number to four significant figures is as follows: 3.002 × 10²
<h3>What are significant figures?</h3>
Significant figures are figures that contribute to the general and overall value of the whole number.
Significant figures or digits are specifically meaningful with respect to the precision of a measurement.
Although, the original number given in this question has 9 significant figures, the number; 300.235800 can be rounded up to four significant figures as follows:
- Decimal notation: 300.2
- No. of significant figures: 4
- No. of decimals: 1
- Scientific notation: 3.002 × 10²
Therefore, the rounding up of the aforementioned number to four significant figures is as follows: 3.002 × 10².
Learn more about significant figures at: brainly.com/question/14359464
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4.) an element cannot be decomposed by a chemical change, it is a pure substance and no other materials compose it.
Answer:
1.23x10^-6 mole
Explanation:
A clear understanding of Avogadro's hypothesis proved that 1mole of any substance contains 6.02x10^23 atoms. This indicates that 1mole of Ag contains 6.02x10^23 atoms.
Now, 1f 1mole of Ag contains 6.02x10^23 atoms, then Xmol of Ag will contain 7.41x10^17 atoms i.e
Xmol of Ag = 7.41x10^17/6.02x10^23 = 1.23x10^-6 mole
Answer:
4) 0.26 atm
Explanation:
In the process:
Benzene(l) → Benzene(g)
ΔG° for this process is:
ΔG° = -RT ln Q
<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>
ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm
1.336 = ln P(benzene) / 1atm
0.26atm = P(benzene)
Right answer is:
<h3>4) 0.26 atm
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Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
