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3 years ago

Heat is which of the following???

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

5 0

A measure of thermal energy transferred between two different bodies at different temperatures would be the correct answer. So, the third option.

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A balloon is filled with 35.0 L of helium in the morning when the temperature is 303 degrees kelvin. By 3:00 p.m., the temperatu

Anna007 [38]

Answer:

The new volume of the balloon is 38.5 L

Explanation:

Step 1: Data given

Volume at the start = V1 = 35.0 L

Temperature at the start = T1 = 303 Kelvin

Volume by 3pm = TO BE DETERMINED

Temperature by 3pm = 333 Kelvin

Step 2: Calculate the new volume

Charles' gas law says

V1/T1 = V2/T2

V 1 is the initial volume and T1 is the initial temperature

V2 is the final volume and T2 is the final temperature

35 L / 303 Kelvin = V2 / 333 Kelvin

V2 = 35L * 333 Kelvin / 303 Kelvin

V2 = 38.47L ≈ 38.5 L

The new volume of the balloon is 38.5 L

7 0

2 years ago

Read 2 more answers

A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5

Marina CMI [18]

Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

1.Use Boyle's Law( $P_{1} V_{1}= P_{2} V_{2}$ ). Re-arrange to solve for $V_{2}$ for the final volume.

2. Plug in values. $V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)} = 7.14 mL$

3 0

1 year ago

Based on the synthesis reaction, what would the product of the reaction be? NaPO3 + CuO → ? NaO + CuPO3 Na + P + Cu + O4 NaCuPO4

trapecia [35]

Answer:

Explanation:

just took test

5 0

3 years ago

Read 2 more answers

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$