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SpyIntel [72]
3 years ago
7

A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

ipitate forms. An excess of lithium sulfate is then added to the solution and a precipitate forms. The precipitate is filtered off and lithium phosphate is added to the remaining solution, producing a precipitate. Which ions are present in the original solution
Chemistry
1 answer:
koban [17]3 years ago
3 0

Answer:

Ca^{2+}  and  Co^{2+}

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, Ca_2 and Co_2

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding Li_2SO_4 to it precipitation takes place.

Precipitate is as follows,

Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)

Thus,

Ca^2^+ is present

When Li_3PO_4 is added again precipitation takes place.

Therefore the reaction is as follows,

Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)

Therefore,

Ca^{2+}  and  Co^{2+} are present in the solution

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Name the particle discovered by James<br> Chadwick in 1932.
LenKa [72]

Answer:

He discovered neutrons in 1932

4 0
3 years ago
A 100. g iron rod is heated to an unknown initial temperature. While cooling down to a temperature of 25.0 degree C, the rod rel
AfilCa [17]

Answer: a)  38.3^0C

Explanation:

As we know that,  

q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

q = heat released  = -600.2 J

m = mass of iron rod =100 g  

T_{final} = final temperature =25.^0C

T_{initial} = initial temperature = ?

c = specific heat of iron = 0.452J/g^0C

Now put all the given values in equation, we get

-600.2J=100\times 0.452J/g^0C\times (25.0-T_{initial})^0C]

T_{initial}=38.3^0C

Therefore, the initial temperature of iron rod was 38.3^0C

6 0
3 years ago
The kilocalories in one can of cola if it has 40. g of carbohydrate and no fat or<br> protein
Marysya12 [62]

Answer:

One can of cola contain 0.16 kcal of carbohydrates.

Explanation:

Given data:

mass of cola can = 40 g

kilocalories of carbohydrates = ?

Solution:

we know that there are 4 cal of carbohydrates per gram.

40 g × 4 cal/g = 160 cal

Now we will convert calories into kilocalories. There are 1000 calories in one kilocalories so,

160/1000 = 0.16 kcal

so, one can of cola contain 0.16 kcal of carbohydrates.

6 0
3 years ago
What are the major specials presents in a solution of a strong acid like HCl
lubasha [3.4K]

Answer:

hydrogen ions (H⁺) and chloride ions (Cl⁻)

Explanation:

Hydrochloric acid (HCl) is a strong acid. That means that the compound dissociates completely into ions when is dissolved in water, as follows:

HCl → H⁺ + Cl⁻

The equilibrium is completely shifted to the right side (products). Thus, it is considered that the concentration of the non-dissociated compound (HCl) is negligible, and the major specials present in the solution are the hydrogen ions (H⁺) and chloride ions (Cl⁻).

4 0
3 years ago
what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield
earnstyle [38]

Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

TiS+H_2O\rightarrow TiO+H_2S

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}

\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g

To calculate the percentage yield of titanium (II) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

5 0
3 years ago
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