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3 years ago

The _________________ moves food from the mouth to the stomach. Once food enters

Chemistry

1 answer:

Tcecarenko [31]3 years ago

7 0

Answer:

oh cool

Explanation:

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The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

tankabanditka [31]

Explanation:

In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.

Hence, the half-cell reaction taking place at anode and cathode will be as follows.

At anode (Oxidation) : $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode (Reduction) : $Ag^{+}(aq) + 1e^{-} \rightarrow Ag(s)$

So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.

$3Ag^{+}(aq) + 3e^{-} \rightarrow 3Ag(s)$ ........ (2)

Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.

$Cr(s) 3Ag^{+}(aq) \rightarrow Cr^{3+}(aq) + 3Ag(s)$

6 0

3 years ago

When an equation is balanced what is the same on both side?

sammy [17]

"equal to"

1=1
1 is equal to 1

Is this what you mean?

3 0

3 years ago

Read 2 more answers

A saline solution similar to that used for intravenous drips is made by dissolving 0.45 g sodium chloride in 50.00 g water. Whic

beks73 [17]

Answer:

E) 1, 2, and 3

Explanation:

50g H2O + 0.45g NaCl --> 50.45g saline solution

7 0

3 years ago

A different solution with an H+ concentration of 1.0 × 10–4 would have a pH = .

tekilochka [14]

1×10^-4 = 0,0001M

pH = -log[H+]
pH = -log0,0001
pH = 4

7 0

3 years ago

What will the concentration of PCl5 be when equilibrium is reestablished after addition of 1.31 g Cl2? PCl5(g) ⇆ PCl3(g) + Cl2(g

Masteriza [31]

Answer:

The new concentration of PCl5 will be 0.01953 M

Explanation:

Step 1: Data given

Mass of Cl2 added = 1.31 grams

Molar mass Cl2 = 70.9 g/mol

Original Equilibrium Mixture:

3.42 g PCl5

4.86 g PCl3

3.59 g Cl2

Volume = 1.0 L

Step 2: The balanced equation

PCl5(g) ⇆ PCl3(g) + Cl2(g)

Step 3: Calculate the original moles and molarity

Moles = mass / molar mass

Moles PCL5 = 3.42 grams / 208.24 g/mol

Moles PCl5 = 0.0164 moles

[PCl5] = 0.0164 M

moles PCl3 = 4.86 grams / 137.33 g/mol

moles PCl3 = 0.0354 moles

[PCl3] = 0.0354 M

moles Cl2 = 3.59 grams / 70.9 g/mol

moles Cl2 = 0.0506 moles

[Cl2] = 0.0506 M

the new mass Cl2 = 3.59 + 1.31 = 4.9 grams

moles Cl2 = 0.0691 moles

[Cl2]= 0.0691 M

The new concentration at the equilibrium

[PCl5] = 0.0164 + X M

[PCl3 ] = 0.0354 - X M

[Cl2] = 0.0691 - X M

Step 4: Calculate Kc

Kc = [Cl2][PCl3] / [PCl5]

Kc = (0.0506*0.0354)/0.0164

Kc = 0.109

Step 5: Calculate [PCl5]

Kc = 0.109 = ((0.0691 - X)(0.0354 - X)) / (0.0164 + X)

X = 0.00313

[PCl5] = 0.0164 + 0.00313 M = 0.01953 M

[PCl3 ] = 0.0354 - 0.00313 M = 0.03227 M

[Cl2] = 0.0691 - 0.00313 M = 0.06597

The new concentration of PCl5 will be 0.01953 M

6 0

3 years ago