Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;
= 0.8
The rate-out
= 
= 
We can say that:
where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12
Integration of the above linear equation =
so we have:
![\frac{d}{dt}[e^{\frac{1}{3}t}A]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5Be%5E%7B%5Cfrac%7B1%7D%7B3%7Dt%7DA%5D)
∴ 
Since A(0) = 12
Then;
Hence;
∴ the concentration at 10 minutes is ;
= 
%
= 0.0456667 %
= 0.046% to three decimal places
The correct answer is A.
B is incorrect because that only applies to nuclear fission.
C is incorrect because it only applies to nuclear fusion.
D is incorrect because energy can be neither created nor destroyed meaning that this statement is physically impossible,
Answer:
[Ca²⁺] = 1M
[NO₃⁻] = 2M
Explanation:
Calcium nitrate dissociates in water as follows:
Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻
The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺
(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺
The concentration of Ca²⁺ is then:
[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M
Similarly, moles of NO₃⁻ can be found using the molar relationship between Ca(NO₃)₂ and NO₃⁻:
(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻
The concentration of NO₃⁻ is then:
[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
Answer:
49.2 g/mol
Explanation:
Let's first take account of what we have and convert them into the correct units.
Volume= 236 mL x (
) = .236 L
Pressure= 740 mm Hg x (
)= 0.97 atm
Temperature= 22C + 273= 295 K
mass= 0.443 g
Molar mass is in grams per mole, or MM= 
or MM= 
. They're all the same.
We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be
n=
, where R (constant)= 0.082 L atm mol-1 K-1
Let's plug in what we know.
n=
n= 0.009 mol
Let's look back at MM= and plug in what we know.
MM= 
MM= 49.2 g/mol
