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tino4ka555 [31]
2 years ago
9

If a gas has a volume of 750 mL at 25°c, what would the volume of the gas be at 55°c?

Chemistry
2 answers:
Airida [17]2 years ago
4 0

825.5mL

To find this, use the equation \frac{V1}{T1} = \frac{V2}{T2}

Make sure to convert all temperatures in Celsius to Kelvin by adding 273. Once you do that, multiply and divide accordingly to get the final answer of 825.5mL

Hope this helps!

Alex2 years ago
3 0

Answer : The final volume of gas will be, 681.4 ml

Explanation :

Charles' Law : This law states that volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T     (At constant pressure and number of moles)

or,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1 = initial volume of gas = 750 ml

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 25^oC=273+25=298K

T_2 = final temperature of gas = 55^oC=273+55=328K

Now put all the given values in the above formula, we get the final volume of gas.

\frac{750ml}{298K}=\frac{V_2}{328K}

V_2=681.4ml

Therefore, the final volume of gas will be, 681.4 ml

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1. Convert each of the following Celsius temperatures to Kelvin. a) 27°C b) 100°C c) 0°C​
valkas [14]

Answer:

a) 300K

b) 373K

c) 273K

Explanation:

to go from °C to K all you have to do is add 273.

5 0
2 years ago
- How many moles of S2 are needed to produce 1.50 moles of SO2 gas? S2 + 20, ----> 2802​
matrenka [14]

Answer:

0.75 moles

Explanation:

S2 + 2O2 = 2SO2

From the reaction above,

We see that number of moles attached to S2 is 1 and number of moles attached to SO2 is 2.

Since we want to find how many moles of S2 are needed to produce 1.50 moles of SO2 gas

The answer is gotten by proportion;

Number of moles = 1/2 × 1.5 = 0.75 moles

3 0
2 years ago
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

6 0
3 years ago
Explain the statement - all bases are not alkalis but all alkalis are bases​
Alexandra [31]

Answer:

A substance that neutralizes an acid is a base and that soluble in water is an alkali. However, all bases are not soluble in water. Thus, All alkali are bases but all bases are not alkali.

Explanation:

hope this helps ✌️

7 0
2 years ago
I need this answer quick please show work
Ainat [17]

Answer:

The answer to your question is 25.2 g of acetic acid.

Explanation:

Data

[Acetic acid] = 0.839 M

Volume = 0.5 L

Molecular weight = 60.05 g/mol

Process

1.- Calculate the number of moles of acetic acid

    Molarity = moles / volume

-Solve for moles

    moles = Molarity x volume

-Substitution

    moles = (0.839)(0.5)

-Result

    moles = 0.4195

2.- Calculate the mass of acetic acid using proportions and cross multiplications

                   60.05 g ----------------------- 1 mol

                        x        ----------------------- 0.4195 moles

                        x = (0.4195 x 60.05) / 1

                        x = 25.19 g

3.- Conclusion

25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M

               

4 0
3 years ago
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