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snow_tiger [21]

4 years ago

8

which weapons did germany used when they had war with Ussr? and which weapons did Ussr used during the war?

1 answer:

MrRissso [65]4 years ago

8 0

Answer: submachine guns

Explanation:first And for most German soldiers took Soviet SSH 39 in ssh40 helmet from Fallen Russians. They were considerably heavier than German models, but provided better Better protection against shrapnel and ammunition from submachine guns, which were widely used by both sides in the war.

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What particles make up the atomic nucleus

White raven [17]

The particles that make up the atomic nucleus of all atoms are both protons and neutrons.

4 0

3 years ago

Read 2 more answers

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

A chemical reaction produces 11.8 moles of tin atoms how many grams of tin are made

valina [46]

Convert mols to grams by multiplying grams of tin by the number of mols.

There are 119 grams per mol

119 x 11.8 = 1404 grams

7 0

3 years ago

Name two compounds in unpolluted air?

goblinko [34]

Nitrogen and oxygen are in unpolluted air

6 0

3 years ago

A chemistry student was asked to calculate the number of moles of iron required to react with 1.20 mol of oxygen to produce iron

Delicious77 [7]

Answer:Write and balance the equation

4Fe + 3O2 -> 2Fe2O3

0.32 mol Fe x 2 mol Fe2O3 / 4 mol Fe =

0.16 mol of Fe2O3

Explanation:

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3 years ago