In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force. ... For the special case of a constant force, the work may be calculated by multiplying the distance times the component of force which acts in the direction of motion.

7 0

3 years ago

What happened to the energy of the system when the baking soda mixed with the vinegar?

andrey2020 [161]

Answer:

Explanation:

I hope it help

It help you mixed up

6 0

3 years ago

A weather balloon at Earth’s surface has a

tigry1 [53]

The temperature : 263.016 K

<h3>Further explanation</h3>

Combined with Boyle's law and Gay Lussac's law

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

P1 = initial gas pressure (N/m² or Pa)

V1 = initial gas volume (m³)

P2 = gas end pressure

V2 = the final volume of gas

T1 = initial gas temperature (K)

T2 = gas end temperature

P1=760 mmHg

V1= 4 L

T1 = 275 K

P2=704 mmHg

V1=4.13 L

$\tt \dfrac{760\times 4}{275}=\dfrac{704\times 4.13}{T_2}\\\\T_2=263.016~K$

6 0

3 years ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

<u>Step 1:</u> Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

<u>Step 2:</u> Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

<u>Step 3:</u> Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

<u>Step 4:</u> Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

4 years ago