Hello!
Antifreeze says it in the name. This means that it helps prevent freezing. The liquid that would freeze if you wouldnt use the antifreeze is water. This means that the antifreeze has to have a higher freezing point then water, because otherwise it would only freeze faster!
Hope this helped you!
Answer:
Intensive properties do not depend on the quantity of matter. Examples include density, state of matter, and temperature. Extensive properties do depend on sample size. Examples include volume, mass, and size.
Explanation:
Brainly!!!
pls
Answer: 0.8M
Explanation:
Given that,
Amount of moles of NaCl (n) = ?
Mass of NaCl in grams = 1.40 g
For molar mass of NaCl, use the molar masses:
Sodium, Na = 23g;
Chlorine, Cl = 35.5g
NaCl = (23g + 35.5g)
= 58.5g/mol
Since, amount of moles = mass in grams / molar mass
n = 1.40g / 58.5g/mol
n = 0.024 mole
Now, given that:
Amount of moles of NaCl (n) = 0.024
Volume of NaCl solution (v) = 30.0mL
[Convert 30.0mL to liters
If 1000 mL = 1L
30.0mL = 30.0/1000 = 0.03L]
Concentration of NaCl solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 0.024 mole / 0.03 L
c = 0.8 M (0.8M means concentration is in moles per litres)
Thus, the concentration of the solution is 0.8M
Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
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
