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taurus [48]
3 years ago
11

Determine the heat needed to warm 25.3 g of copper from 22 degrees celsius to 39 degrees celsius.

Chemistry
1 answer:
Serggg [28]3 years ago
5 0

Answer:

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

Explanation:

Q=mc\Delta T

Where:

Q = heat absorbed  or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have mass of copper = m = 25.3 g

Specific heat of copper = c = 0.385 J/g°C

ΔT  = 39°C - 22°C = 17°C

Heat absorbed by the copper :

Q=25.3 g\times 0.385 J/g^oC\times 17^oC=165.59 J

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

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What is Cd2+ and F1- together as a formula
geniusboy [140]

Answer:

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Explanation:

Cd ²+ + F¹- = CdF2

F fluor is a non metal so he takes the electons that Cd gives so if Vd has to give 2 electrons and F can take only 1, there has to be 2 F atoms so all the electrona can be neutral and in ionisation

4 0
3 years ago
Use the van der Waals equation and the ideal gas equation to calculate the pressure for 1.50 mol He gas in a 1.00 L container at
Tamiku [17]

The difference in pressure is <u>8.98 %.</u>

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3 0
1 year ago
Fill in the blanks to complete each statement about the heating of Earth's surface
Ad libitum [116K]
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6 0
3 years ago
Read 2 more answers
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
4 years ago
What is the temperature of 0.500 moles of a gas that is stored in a 2.50 L container under 13.0 atm of pressure?
faust18 [17]
<h2>Answer:</h2>

Moles of a gas = 0.500

Volume = 2.50 L

Pressure = 13. atm

Temperature = ?

Solution:

Formula:

                       PV = n RT

Putting the values in formula:

 T = PV/nR = 13 * 2.5 / 0.5 *  0.082057

   =  32.5/0.041 = 792.68 K

                    T = 792.68 K

3 0
4 years ago
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