Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law: 
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means 
and 
Lastly, we must calculate the number of moles of 
there are. Given 0.80g of , we will need to convert with the molar mass of . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: 
Thus, 
Isolating V in the Ideal Gas Law:
...substituting the known values, and simplifying...
So, 0.80g of would occupy 0.56L at STP.
Isoleucine' is the chemical name of 'titin' (also known as 'connectin') - the largest known protein. It has 189,819 letters.
<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".
V </span>α T
Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂
V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K
By applying the formula,
2.00 L / 313 K = V₂ / 303 K
V₂ = (2.00 L / 313 K) x 303 K
V₂ = 1.94 L
Hence, the volume of the balloon at 30.0 ⁰C is 1.94 L
Given:
175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)
The balanced chemical equation is shown below:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.
Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
mass of NH3 = 185.94 kg NH3 needed
mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg
mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O
