Which molecule is a stereoisomer of trans-2-butene?

-Dominant- [34]

According to the task, you are proveded with patial pressure of CO2 and graphite, and here is complete solution for the task :
At first you have to find n1 =moles of CO2 and n2 which are moles of C
<span>The you go :
</span> $CO2(g) + C(s) \ \textless \ =\ \textgreater \ 2CO(g)$

n1 n2 0
-x -x +2x
$n1-x n2-x 2x Kp=P^2 (CO)/Pco2$
After that you have to use the formula $PV=nRT$
$Kp=(2x)^2/n1-x$
Then you have to solve x, and for that you have to use <span>RT/V
And to find total values:</span> $P(total)=n(total)*R*T/V$
I am absolutely sure that this would be helpful for you.

7 0

3 years ago

List the methods of separation used in water purification...... ( HELP PLEASE )

lawyer [7]

Answer:

I can list four. These are the main ones.

Bolling

Filtration

Distillation.

Chlorination

Explanation:

3 0

3 years ago

SCIENCE:how can i model a ice cream maker melting!?

lord [1]

Answer:

BURN IT ALIVE MUHAHAHAHAHA

Explanation:

jk

6 0

2 years ago

Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a

GrogVix [38]

Answer:

The calorie content of 65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium) + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41 * (65/15) = 326.78 cal

8 0

2 years ago

Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio

grigory [225]

Answer : The equilibrium concentration of $CH_3OH$ will be, (C) $2.82\times 10^{-1}$

Explanation : Given,

Equilibrium constant = 14.5

Concentration of $CO$ at equilibrium = 0.15 M

Concentration of $H_2$ at equilibrium = 0.36 M

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get:

$14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}$

$[CH_3OH]=2.82\times 10^{-1}M$

Therefore, the equilibrium concentration of $CH_3OH$ will be, (C) $2.82\times 10^{-1}$

5 0

3 years ago