Answer:
Mg
Explanation:
The standard reduction potentials are
<u>E°/V
</u>
Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42
Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85
Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Mg2+(aq) + 2e- ⟶ Mg(s); -2.38
The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
Mg is the only metal with a standard reduction potential lower than that of Cu, so
Only Mg will react spontaneously with Cu²⁺.
A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²
B. The cost of the patch to the nearest cent is 734 cents
<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>
We can convert 16.1 cm² to km² as illustrated below:
Conversion scale
1 cm² = 1×10⁻¹⁰ km²
Therefore,
16.1 cm² = 16.1 × 1×10⁻¹⁰
16.1 cm² = 1.61×10⁻⁹ km²
Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²
<h3>B. How to determine the cost in cent</h3>
We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:
1 cm² = 0.155 in²
Therefore,
16.1 cm² = 16.1 × 0.155
16.1 cm² = 2.4955 in²
Finally, we shall the determine the cost in centas fo r llow:
- Cost per in² = $2.94 = 294 cent
- Cost of 2.4955 in² =?
1 in² = 294 cent
Therefore,
2.4955 in² = 2.4955 × 294
2.4955 in² = 734 cents
Thus, the cost of the patch is 734 cents
Learn more about conversion:
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Now ,
C + O2 → CO2
According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.
No of moles = mass of the substance/molecular mass of the substance.
In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.
No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.
No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles
Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.
Also number of moles of O2 = mass of O2÷ molar mass of O2
Substituting number of moles of O2 as 0.1 we get
mass of O2(x) = Number of moles of O2 × Molar mass of O2
Mass of O2 (x) = 0.1 × 32= 3.2 g
Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.
