All categories

2 years ago

According to reference table adv-10, which reaction will take place spontaneously?

1 answer:

5 0

Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V

<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V

V= - 0.777V (no spontaneous)

You might be interested in

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

3 0

2 years ago

Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Pepsi [2]

Answer:

$Mass_{chromium}=52.1\ g$

$Mass_{neon}=8.23\times 10^{-7}\ g$

Explanation:

<u>Calculation of the mass of chromium as:- </u>

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1.002\ mol= \frac{Mass}{51.9961\ g/mol}$

$Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g$

<u>Calculation of the mass of neon as:- </u>

Moles = $4.08\times 10^{-8}$ moles

Molar mass of neon = 20.1797 g/mol

Thus,

$1.002\ mol= \frac{Mass}{20.1797\ g/mol}$

$Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g$

6 0

2 years ago

Jamal wants to make a model of a hill near his house to test the way the slope affects how rain runs down the hill. Which type o

Nikitich [7]

a scale-model mound made of the same materials that make the real hill

4 0

2 years ago

Read 2 more answers

Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4

aleksandr82 [10.1K]

Answer:

Molarity for the sulfuric acid is 0.622 M

Explanation:

When we neutralize an acid with a base, molarity of both . both volume are the same. The formula is:

M acid . volume of acid = M base . volume of base

M acid = unknown

Volume of acid = 17 mL

Volume of base = 45 mL

M base = 0.235 M

Therefore, we replace: M acid . 17 mL = 0.235 M . 45 mL

M acid = (0.235 M . 45 mL) / 17 mL

M acid = 0.622 M

6 0

2 years ago

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

Harlamova29_29 [7]

Answer: Molarity of $MgCl_2$ in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{\text{no of moles}}{\text{Volume in L}}$

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.96}{1.00L}$

$Molarity=1.96mol/L$

Thus molarity of $MgCl_2$ in the original sample was 1.96M

4 0

2 years ago