The pressure of the CO₂ = 0.995 atm
<h3>Further explanation</h3>
The complete question
<em>A student is doing experiments with CO2(g). Originally, a sample of gas is in a rigid container at 299K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425K.</em>
<em>Calculate the pressure of the CO₂ (g) in the container at 425 K.</em>
<em />
<em />
Gay Lussac's Law
When the volume is not changed, the gas pressure is proportional to its absolute temperature
P₁=0.7 atm
T₁=299 K
T₂=425 K
<em />
It's an example of a molecule
<span>To find the molar mass, look at a periodic table for each element.
Ibuprofen, C13 H18 and O2. Carbon has a molar mass of 12.01 g, Hydrogen has 1.008 g per mole, and Oxygen is 16.00 g per mole.
C: 13 * 12.01
H: 18 * 1.008
O: 2 * 16.00
Calculate that, add them all together, and that is the molar mass of C13H18O2.
Molar mass: 206.274
Next, you have 200mg in each tablet, with a ratio of C13H18O2 (molar mass) in GRAMS per Mole
So, you need to convert miligrams into grams, which is 200 divided by 1000.
0.2 g / Unknown mole = 206.274 g / 1 Mole
This is a cross multiplying ratio where you're going to solve for the unknown moles of grams per tablet compared to the moles per ibuprofen.
So, it's set up as:
0.2 g * 1 mole = 206.274 * x
0.2 = 206.274x
divide each side by 206.274 to get X alone
X = 0.00097
or 9.7 * 10^-4 moles
The last problem should be easy to figure out now that you have the numbers. 1 dose is 2 tablets, which is the moles we just calculated above, times four for the dosage.
</span>
Answer:
Aluminium Chloride + Hydrogen
Please vote for Brainliest and I hope this helps!
Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:
w is the mass flow
m is the mass of salt
v is the volume flow
C is the concentration
![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
