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Sav [38]
2 years ago
13

True or false? To decrease the concentration of a solution, add more liquid.

Chemistry
1 answer:
Setler79 [48]2 years ago
8 0
It’s true !!!!!!!!!!!!!!!! N
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What is the graph of the orange line and is it a positive or negative connotation
dimulka [17.4K]

Answer:

Positive

Explanation:

4 0
2 years ago
A reaction that occurs in the internal combustion engine is N₂(g) + O₂(g) ⇄ 2NO(g) (c) What is the significance of the different
vodomira [7]

The response would become spontaneous if the value of  ΔG° was negative.

According to the estimated value of  ΔG°, it is shown that  ΔG° value decreases as temperature value increases. The value shifts from being more favorable to being less favorable. It would appear that the value of  ΔG° would be negative at a specific temperature, causing the reaction to occur spontaneously.

The reaction is in an equilibrium state if  ΔG = 0. If ΔG < 0, the reaction is spontaneous in the direction written. The relationship between terms from the equilibrium is paralleled by the relevance of the sign of a change in the Gibbs free energy.

Learn more about ΔG° here:

brainly.com/question/14512088

#SPJ4

8 0
1 year ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is
Gnoma [55]
3.heat of reaction 
hope that helped
3 0
3 years ago
You are a NASCAR pit crew member. Your employer is leading the race with 20 laps to go. He just finished a pit stop and has 5.0
jek_recluse [69]

Answer:

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

Explanation:

Density = mass / volume

Density of fuel = 700 g/ 1 gal

Therefore, the mass of fuel in 1 gallon = 700 g

The driver has 5.0 gallons of fuel in the tank.

The mass of 5.0 gallons of fuel = 5 × 700 = 3500 g of fuel

Equation of the combustion of fuel, C₅H₁₂ is given below:

C₅H₁₂ + 8 O₂ ---> 6 H₂O + 5 CO₂

1 mole C₅H₁₂ requires 8 moles of O₂

1 mole of C₅H₁₂ has a mass = 72 g

8 moles of O₂ has a mass = 256 g

Therefore, 300 g of O₂ will require 300 × (72/256) g of C₅H₁₂ = 84.375 g of C₅H₁₂

84.375 g of fuel is used by the car per lap;

20 laps will require 20 × 84.375 g of fuel = 1687.5 g of fuel.

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

7 0
3 years ago
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