Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
T=20 min
m₀=200 g
t=60 min
the mass of element through time t is:
m=m₀*2^(-t/T)
m=200*2^(-60/20)=25 g
25 grams of element will be left after 60 minutes
Answer:
The mass of one mole of a substance is equal to that substance's molecular weight. ... water is 18.015 atomic mass units (amu), so one mole of water weight 18.015 grams. ... Avogadro's number is a proportion that relates molar mass on an atomic ... one molecule of water (H2O), one mole of oxygen (6.022×1023 of O atoms)
Answer:
The correct answer is option B: "I, II".
Explanation:
The synthesis of 5-phosphoribosylamine from phosphoribosyl pyrophosphate (PRPP) is inhibited allosterically by ATP and GTP and activated by PRPP via feed forward activation. The enzyme that regulate this process is the enzyme amidophosphoribosyltransferase (AMPRT), which is stimulated by increased PRPP concentrations and is inhibited allosterically by ATP and GTP as well as IMP, AMP, and GMP.
Option B. <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO is the correct answer. HOPE IT HELPS</span>
