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stiv31 [10]
3 years ago
12

How would you solve this?

Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
6 0
I’m sorry but I did not understand the question
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3. Dorji was provided with four colourless solutions. He was asked to
givi [52]

Answer:

First, we can test Solution 1. We know that Sodium Hydroxide is a strong base. If we test acids on blue litmus paper, they will turn red. If we test bases on red litmus paper, they will turn blue.  So, you can test all the of the solutions- water, sodium hydroxide and hydrochloric acid with blue and red litmus paper. HCl, Hydrochloric acid is an acid, so it will turn blue litmus paper red. It will not turn red litmus blue. The acids will turn blue litmus paper red. The bases will turn red litmus paper blue. Only water is a neutral liquid, which will not turn blue litmus paper red or red litmus paper blue. It will not change the colour of it. Thus, if you test all the solutions with blue and red litmus paper, you will know which solution is water.  Water is the only one which is neutral. It is the only solution which cannot change the colour of any litmus paper. Thus, you can identify it very easily.  

5 0
3 years ago
Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper
KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

  • CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0
3 years ago
Why is it important for organelles to double during the G2 phase of the cell cycle?
boyakko [2]

Explanation:

so then we would know about the reproduction system and how we are born and what happens with the cells inside the womens womb and how the sperm cell meets the egg cell and creates the egg

6 0
3 years ago
A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet
stira [4]

Explanation:

The given data is as follows.

           Mass of apple sauce mixture = 454 kg

           Heat added (Q) = 121300 kJ

 Heat capacity (C_{p}) of apple sauce at 32.8^{o}C = 4.0177 kJ/kg^{o}C

So, Heat given by heat exchanger = heat taken by apple sauce

                            Q = mC_{p} \Delta T

or,                    Q = mC_{p} (T_{f} - T_{i})  

Putting the given values into the above formula as follows.

                     Q = mC_{p} (T_{f} - T_{i})  

              121300 kJ = 454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)

                      T_{f} = 76.5^{o}C

Thus, we can conclude that outlet temperature of the apple sauce is 76.5^{o}C.

3 0
4 years ago
Which characteristic is a property of water?
Anettt [7]
D.
The main properties of water are its polarity, cohesion, adhesion, surface tension, high specific heat, and evaporative cooling.
7 0
3 years ago
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