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daser333 [38]
3 years ago
9

Steps in complete balanced equation of a process

Chemistry
1 answer:
Vlad [161]3 years ago
7 0

Answer:

Step 1: The Unbalanced Chemical Equation. The unbalanced chemical equation is given to you.

Step 2: Make a List.

Step 3: Identifying the Atoms in Each Element.

Step 4: Multiplying the Number of Atoms.

Step 5: Placing Coefficients in Front of Molecules.

Step 6: Check Equation.

Step 7: Balanced Chemical Equation.

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Calculate the Kelvin temperature to which 10.0 L of a gas at 27 °C would have to be heated to change the volume to 12.0 L. Units
KatRina [158]

Hello!

For this problem, we will be applying <em>Charles' Law</em>:

V1/T1 = V2/T2

Now that we have the formula, let's convert the temperature to Kelvin.

27 + 273 = 300K

Let's plug everything in now!

10/300 = 12.0/x

Simplified:

1/30 = 12.0/x

Cross-multiply:

1x = 30*12.0

<u>x = 360</u>

<em>Check!</em>

10/300 = 12/360

300*12 = 360*10

3600 = 3600

Therefore, you would have to heat the gas at a temperature of 360K in order to raise the volume to 12.0L.

 

4 0
3 years ago
What is the definition of an ion.
mario62 [17]

<u><em>*(Answer)*=</em></u> An ion can have a positive or negative charge. Furthermore, it is a an atom or molecule with a net electric charge due to the loss or gain of one or more electrons.

<u><em>*(Request)*=</em></u> Searching the definition can help you find a definition faster.

Hope this helps

User who Answered: <u><em>BangtanBoyScouts</em></u>

4 0
4 years ago
Why is cl2 a subcript​
vazorg [7]

Answer:

In Cl_2, the 2 is a subscript because it indicates there are 2 of the same elements. The Lewis structure would display it as Cl-Cl.

On the other hand, a superscript would indicate a specific charge.

All subscripts show the amount of the specific element there is.

An example would be O_2\\ or N_2, they both show that there are 2 of the same elements.

If the subscript is outside a parenthesis such as Ca_3(PO_4)_2 it indicates there are 2 PO_4 molecules.

5 0
3 years ago
If 200 grams of water is to be heated from 24.0oC to 100.0oC to make a cup of tea, how much heat must be added?
aksik [14]
According to this formula:

Q = M*C*ΔT 

when we have M ( the mass of water) = 200 g 

and C ( specific heat capacity ) of water = 4.18 J/gC

ΔT (the difference in temperature) = Tf - Ti 
     
                                                          = 100 - 24

                                                          = 76°C

So by substitution:


Q = 200 g * 4.18 J/gC * 76 °C

   = 63536 J 

 ∴ the amount of heat which be added and absorbed to raise the temp from 24°C to 100°C is = 63536 J


3 0
4 years ago
Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2
quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done

p_{ext} = external pressure

V_1 = initial volume of gas

V_2 = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

w=-nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 25^oC=273+25=298K

V_1 = initial volume of gas

V_2 = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(2.5atm)\times (10-1)L

w=-22.5L.atm=-22.5\times 101.3J=-2279.25J

(2) A free isothermal expansion from 1 L to 100 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})

w=-11409.6J

(3) A reversible isothermal expansion from 0.5 L to 4 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})

w=-5151.97J

(4) A reversible isothermal expansion from 0.5 L to 40 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})

w=-10856.8J

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(0.5atm)\times (100-1)L

w=-49.5L.atm=-49.5\times 101.3J=-5014.35J

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0
3 years ago
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