Write the products and balance the equation:<br><br> C4H7OH + O2 → CO2 +H2O

A balloon has a volume of 6.2 liters at 23.2 C. The balloon is then heated to a temperature of 144.0 C. What is the volume of th

Marina CMI [18]

Answer:

8.7 L

Explanation:

T2(V1/T1) = V2

417.15 K(6.2 L/296.45 K) = 8.7 L

Remember to almost always change celcius to kelvin. Also, this is part of Charle's Law (temp and volume are proportional, so if temp increaces so must the volume or vice versa). Lastly, Charle's Law has the formula of V1/T1 = V2/T2. I just rearranged it to go along with your problem. Hence, the T2(V1/T1) = V2

4 0

3 years ago

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/m

koban [17]

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;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

3 years ago

Text me if im ur type im looking for a girlfreind

jasenka [17]

Answer:

Argon is an element on the periodic table. It is a member of the noble gasses, meaning that it is a monoatomic pure gas. In other words, argon gas particles are simply pure single atoms of argon. ... Finally, water is a polyatomic molecule H2O composed of two atoms of hydrogen and one atom of oxygen.

Explanation:

3 0

3 years ago

Read 2 more answers

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0

3 years ago

15

Fynjy0 [20]

Answer:

larva hatching, larva with legs, young newt, adult newt, in that order

4 0

2 years ago

Write the products and balance the equation: C4H7OH + O2 → CO2 +H2O