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sasho [114]
3 years ago
14

Which Salt is least soluble at 50 C

Physics
1 answer:
Anvisha [2.4K]3 years ago
4 0

The salt c e_{2}\left(s o_{4}\right)_{3} is the least soluble salt at 50^{\circ} \mathrm{C}.

Answer: Option 1

<u>Explanation: </u>

As it is clear that c e_{2}\left(s o_{4}\right)_{3}, a gas so the solubility of it decreases on increasing the temperature of water. Because gas molecule starts to evaporate on attaining heat with the increase of temperature thus leading to reduction in the solubility of them.

As most of the gas molecules get vaporised the solubility in water gradually decreases for this compound. It is also clear from the solubility graphs of the various salts.

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Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
4 years ago
D
inysia [295]

<u>Hello and Good Morning/Afternoon</u>:

<em>Original Question: C₂H₅OH + __O₂ → __CO₂ + __ H₂O</em>

<u>To balance this equation</u>:

⇒ must ensure that there is an equal number of elements on both sides of the equation at all times

<u>Let's start balancing:</u>

  • On the left side of the equation, there are 2 carbon molecule

              ⇒ but only so far one on the right side

         C<em>₂H₅OH + __O₂ →  2CO₂ + __ H₂O</em>

  • On the left side of the equation, there are 6 hydrogen molecules

               ⇒ but only so far two on the right side

         C<em>₂H₅OH + __O₂ →  2CO₂ + 3H₂O</em>

  • On the right side of the equation, there are 7 oxygen molecules

                ⇒ but only so far three on the left side

         C<em>₂H₅OH + 3O₂ →  2CO₂ + 3H₂O</em>

<u>Let's check and make sure we got the answer:</u>

                           C<em>₂H₅OH + 3O₂ →  2CO₂ + 3H₂O</em>

<em>                 2 Carbon                ⇔                    2 Carbon</em>

<em>                 6 Hydrogen            ⇔                   6 Hydrogen</em>

<em>                 7 Oxygen                ⇔                   7 oxygen</em>

<u>Thefore the coefficients in order are</u>:

  ⇒ 1, 3, 2, 3

<u>Answer: 1,3,2,3</u>

Hope that helps!

#LearnwithBrainly<em>                      </em>

5 0
2 years ago
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