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3 years ago

9

If the cooling power remains constant, what will be the temperature of the system TB after it has been in the freezer for exactl

y 1 hour

1 answer:

serg [7]3 years ago

8 0

Answer:

The answer is "-50.424° C".

Explanation:

Given:

Time = 1 hour

TB=?

Formula:

$TB= \frac{-T\times P}{m \times c_i}$

time (T) = 1 hour

= 3600 second

$t=(3600 -t_2)$

Calculating T_B:

$\to T_B= \frac{-(3600 - t_2) 36.6}{0.250 \times 2090}\\$

$= \frac{-(3600 - 2880) 36.6}{522.6}\\\\= \frac{-(720) 36.6}{522.6}\\\\= \frac{-720 \times 36.6}{522.6}\\\\= \frac{-26352}{522.6}\\\\= -50.424 ^{\circ} C$

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For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in

Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>

<em>C(s) + 2H2(g) <--> CH4(g). </em>

<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

<u><em>PpCH4 = 1.44 atm</em></u>

6 0

3 years ago

PLZ HELP ITS DUE IN AN HOUR I WILL GIVE BRAINLISTS TO THE RIGHT ANSWER

VARVARA [1.3K]

Answer:

Explanation:88

6 0

2 years ago

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago

A sample of gas has a volume of 20.0 liters at 22.0° C. If the pressure remains constant, what is the volume at 100.0° C?

miv72 [106K]

The problem applies Charles' law since constant pressure with varying volume and temperature are given. Assuming ideal gas law, the equation to be used is $\frac{ V_{1} }{ T_{1} }$ = $\frac{ V_{2} }{ T_{2} }$ . We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.

6 0

3 years ago

Read 2 more answers

What are the concentrations of h3o+ and oh− in tomatoes that have a ph of 4.10?

IRINA_888 [86]

Answer : The concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

Solution : Given,

pH = 4.10

pH : pH is defined as the negative logarithm of hydronium ion concentration.

Formula used : $pH=-log[H_3O^+]$

First we have to calculate the hydronium ion concentration by using pH formula.

$4.10=-log[H_3O^+]$

$[H_3O^+]=antilog(-4.10)$

$[H_3O^+]=7.94\times 10^{-5}$

Now we have to calculate the pOH.

As we know, $pH+pOH=14$

$4.10+pOH=14$

$pOH=9.9$

Now we have to calculate the hydroxide ion concentration.

$pOH=-log[OH^-]$

$9.9=-log[OH^-]$

$[OH^-]=antilog(-9.9)$

$[OH^-]=1.258\times 10^{-10}$

Therefore, the concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

5 0

2 years ago