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iogann1982 [59]
2 years ago
14

HELP NEED THIS DONE BEFORE 9:45!!!!!!!!!!!!!!!! WILL GIVE 100 POINTS AND BRAILIEST IF DONE BEFORE 9:45!!!!!!!!!!!!!

Chemistry
2 answers:
ehidna [41]2 years ago
7 0

Answer:

1.) 742.57 m/s

2.) 132

3.) 4.375 hours

4.) 1080

5.) 81.6 km per hour

6.) 20 min = 1,200 secs1000 / 1200

Explanation:

Your Welcome!

sveticcg [70]2 years ago
7 0

Answer:

1.) 742.57 m/s

2.) 132 m

3.) 4.375 hours

4.) 1080 m

5.) 81.6 km per hour

6.) 20 min = 1,200 secs1000 / 1200

Explanation:

**Brainliest would be appreciated**

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A galvanic cell is formed when two metals are immersed in solutions differing in concentration 1 when two different metals are immersed.

<h3>What is galvanic cell?</h3>

A galvanic cell is an electrochemical device that transforms chemically generated free energy into electrical energy. A photogalvanic cell produces photochemical species that react to produce an electrical current when connected to an external circuit.

<h3>How does galvanic cell works?</h3>

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2 years ago
Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under t
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Answer:

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Explanation:

Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:

Q = c * m * ΔT

Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).

In this case:

  • c=4.18 \frac{J}{g*K}
  • m= 1.30 kg= 1,300 g (1 kg=1,000 g)
  • ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K  Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.

Replacing:

Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K

Q= 64,121.2 J

<u><em>The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J</em></u>

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3 years ago
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