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3 years ago

Please help! 20 Points!!!

Chemistry

2 answers:

-BARSIC- [3]3 years ago

8 0

Answer:

ag+ and zn2+

Explanation:

second question is A

lorasvet [3.4K]3 years ago

7 0

Answer:

I just did it home slice the first one's Ag+ and Zn2+ and the second one is A

Explanation:

I just did the assignment

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What are did facts about atomic number 20?

stira [4]

<span></span>Basic Calcium Facts

Name: Calcium

Atomic Number: 20

Element Symbol: Ca

Group: 2

Period: 4

Block: s

Element Family: Alkaline Earth

Atomic Mass: 40.078(4)

Electron Configuration: [Ar]4s<span>2
</span>

Full: 1s22s22p63s23p64s2 (full)

3 0

3 years ago

What volume of chlorine gas at 27 °C, 812 mmHg, is required to react with an excess of carbon disulfide so that 5.00kJ of heat i

Kamila [148]

Answer:

The correct answer is 1.21 L.

Explanation:

Based on the given information, the reaction will be,

CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)

By using the standard values of the substances, the standard enthalpy of the reaction is,

ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol

= -285.3 kJ/mol

The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.

Now the number of moles of chlorine needed to react to produce 5.00 kJ is,

= 5.00 kJ × 3 mol Cl2/285.3 kJ

= 0.0526 mol Cl2

Now the volume of chlorine gas at 27degree C and 812 mmHg will be,

Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm

= 1.21 L

4 0

3 years ago

A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0

4 years ago

What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention

inysia [295]

The name of Pb(NO3)2 of is Lead II trioxonitrate V

The bond which exist in is Lead II trioxonitrate V ( Pb(NO3)2 ) are both ionic and covalent bond

<h3>What is a compound?</h3>

A compound is a substance which contains two or more elements chemically combined together

So therefore, the name of Pb(NO3)2 of is Lead II trioxonitrate V

Learn more about a chemical compound:

brainly.com/question/26487468

#SPJ1

4 0

2 years ago

an alloy composed of tin, lead, and cadmium is analyzed. the mole ratio of sn:pb is 2.73:1.00, and the mass ratio of pb:cd is 1.

nordsb [41]

This problem is describe the mole-ratio composition of an allow composed by tin, lead and cadmium. Ratios are given as Sn:Pb 2.73:1.00 and Pb:Cd is 1.78:1.00, and we are asked to calculate the mass percent compositon of Pb in the allow.

In this case, according to the given information, it turns out possible realize that the following number of moles are present in the alloy, according to the aforementioned ratios:

$2.73mol Sn\\\\1.00molPb\\\\\frac{1.00molPb*1.00molCd}{1.78molPb}= 0.562molCd$

Next, we calculate the masses by using each metal's atomic mass:

$m_{Sn}=2.73mol*\frac{118.7g}{1mol}=324.05g\\\\ m_{Pb}=1.00mol*\frac{207.2g}{1mol}=207.2g\\\\m_{Cd}=0.562mol*\frac{112.4g}{1mol}=63.2g$

Thus, the mass percent composition of each metal is shown below:

$\%Sn=\frac{324.05g}{324.05g+207.2g+63.2g} *100\%=54.5\%\\\\\%Pb=\frac{207.2g}{324.05g+207.2g+63.2g} *100\%=34.9\%\\\\\%Cd=\frac{63.2}{324.05g+207.2g+63.2g} *100\%=10.6\%$

So that of lead is 34.9 %.

Learn more:

8 0

3 years ago