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zubka84 [21]
3 years ago
7

What is 1 +1 (a) 11 (b) 3 (c) 6 (d) 2

Chemistry
2 answers:
Doss [256]3 years ago
8 0
I believe the answer is D) 2
kodGreya [7K]3 years ago
3 0

Answer:

(d) 2

Explanation:

Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer

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Nat2105 [25]

A good reason for a desert fox to show this pattern of behavior because hunting at night allows the fox to use its night vision.

<h3>What is Hunting?</h3>

Thi9s is commonly practised by predators such as fox in which they capture and kill other animals for food.

The fox has a good night vision which makes it able to hunt for animals during the night also. This is why option C is chosen as the most appropriate choice.

Read more about Hunting here  brainly.com/question/81175

6 0
2 years ago
What is determined by calculating the slope of the position vs time graph?
vova2212 [387]

Answer: Velocity

Explanation:

4 0
3 years ago
Read 2 more answers
Need help on question 3 plz<br> URGENT HELP PLZ
Law Incorporation [45]
Alloys are the homogeneous mixture of metals and non metals.... they are used in making cars and jewellery to make them more featurable,,, means that they would possess high regidity to the destruction .... both features of metal and nonmetal
5 0
3 years ago
What is the number of cations is present in 1.17 g of sodium chloride? Given Na=6*10power 23 mol
natita [175]
1.17×6×1023 Wich equals 7.02×1023 Your final answer is 7.02×1023
5 0
3 years ago
Read 2 more answers
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
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