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san4es73 [151]
3 years ago
14

V. = 22.4 L; P= 1 atm; P, = ? atm; V, = 2.8 L

Chemistry
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

The pressure is 8 atm

Explanation:

We use the Boyle Mariotte formula, where for a given mass of gas at constant temperature, the pressure and volume vary inversely:

P1xV1 = P2xV2

P2= (P1xV1)/V2= (1 atm x 22,4 L)/ 2,8L= 8 atm

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amm1812

What do you mean bro

7 0
3 years ago
a mixture containing 21.4 g of ice. (0.00 c) and 75.3 g of water (55.3 c) is placed in an insulatated container
Leto [7]
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7 0
4 years ago
A question on Reactants.
Natasha2012 [34]

Answer:

b lithium and water

Explanation:

8 0
3 years ago
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
How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

6 0
3 years ago
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