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sleet_krkn [62]
3 years ago
6

Calculate the solubility of silver chloride in 0.17M AgNO3. (Ksp=1.8E-10)

Chemistry
1 answer:
Zina [86]3 years ago
4 0
  As the given equation is
AgNO3 = Ag+ + NO3-
split it then derive the
 the Ksp expression.

Ksp = [Ag+] [NO3-]

Since there is 1:1 molar ratio,so by applying values

<span>1.8E^-10 = (s)(s)
</span><span>1.8E^-10 = s^2
</span><span>s=1.34 x 10^-5 M
this is the solubility of silver chloride
hope it helps</span>
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predict where an unknown element that has these properties could fit in the periodic table 62 protons and electrons
Vesnalui [34]
Position of element in periodic table is depend on the electronic configuration of element.

Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.

Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.

Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table. 
5 0
3 years ago
What change in mass number occurs when a radioactive atom emits an alpha particle?
Alik [6]
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6 0
3 years ago
Glven 4.62 mol of sodium hydroxide, NaOH, how many grams are present?
Vaselesa [24]
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Hope this helped
3 0
3 years ago
Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr
leonid [27]
The law that states the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pressure conditions would be the Charles Law. It <span>is an experimental gas </span>law<span> that describes how gases tend to expand when heated. Hope this answers the question.</span>
6 0
3 years ago
The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a
stealth61 [152]

<u>Answer:</u> The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

<u>Explanation:</u>

We are given a function that calculates the amount of sample remaining after 't' years, which is:

A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}

  • <u>For t = 20 years</u>

Putting values in above equation:

A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}

A_t(t)=288.522g

Hence, the amount of sample left after 20 years is 288.522 g

  • <u>For t = 50 years</u>

Putting values in above equation:

A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}

A_t(t)=144.26g

Hence, the amount of sample left after 50 years is 144.26 g

6 0
3 years ago
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