The new volume of the hydrogen gas in the balloon is 1.5 L.
<u>Explanation:</u>
At STP, Temperature, T1 is 0° C = 0 + 273 K = 273 K
Pressure, P1 = 1 atm
Here given volume, V1 = 3.60 L
Balloon at sea,
Pressure, P2 = 2.5 atm
Temperature, T2 = 10° C = 10 + 273 = 283 K
Volume, V2 = ?
Here, we have to use the equation,

We have to rearrange the equation to get V2 as,

Now plugin the values as,
V2 = 
= 1.5 L
So the new volume of the hydrogen gas is 1.5 L.
1.679 × 10²³ g of PbO₂ is present in 4.23×10⁴⁴ particles of PbO₂.
<u>Explanation:</u>
First the number of particles is converted into moles by dividing it by the Avogadro's number and then moles multiplied by the molar mass, we will get the mass of PbO₂ in grams.
4.23×10⁴⁴/ 6.022×10²³ = 7.02×10²⁰ moles
Now we have to multiply this by the molar mass of PbO₂, that is 239.2 g/mol, we will get the mass in grams.
7.02×10²⁰ moles × 239.2 g/mol = 1.679 × 10²³ g of PbO₂
Chemical change is most often indicated by a change in color.
In accordance with Dalton's Law of multiple proportions
<h3>Further explanation</h3>
Given
6.0g of carbon
22.0g or 14.0g of product
Required
related laws
Solution
the amount of air present ⇒ as an excess or limiting reactant
- air(O₂) as a limiting reactant(product=14 g)
C+0.5O₂⇒CO
6 + 8 = 14 g
mol O₂=8 g : 32 g/mol=0.25
mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)
mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g
- air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant
C+O₂⇒CO₂
6 + 16 = 22 g
mol C = 6 g : 12 g/mol = 0.5
mol O₂ = 16 g : 32 g/mol=0.5
mol CO₂ = 22 g : 44 g/mol = 0.5
if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.
CO = 6 : 8
CO₂ = 6 : 16
the ratio O = 8 : 16 = 1 : 2
In accordance with Dalton's Law of multiple proportions
Answer:
C. grazing, mining, developing oil and gas.
Explanation:
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Hope it helps.