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3 years ago

You need to separate two substances that make up a compund hoe coukld u go about doing it

1 answer:

5 0

Compounds are chemically separated unlike mixtures. To separate one you could burn it or dissolve it in something to make the compound separate into its two elements.

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How can i prove that nascent oxygen more reactive than normal oxygen ?

laila [671]

Nascent oxygen has much higher reactivity than the oxygen bubbled through the reaction mixture. It doesn't stay nascent for long (you are right about it being converted quick to just O2), which is why it has to be generated in situ

6 0

3 years ago

What is the specific heat of copper when heated to 221.32

Dafna1 [17]

Answer:

The specific heat of copper when heated to 221.32 (not listed form of heat measurement) is 221.32 (not listed form of heat measurement).

Explanation:

uh not really sure what else there is here, I may be missing something

5 0

2 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

6 0

3 years ago

Determine the oxidation number of the element "J" in H3JO2-

igor_vitrenko [27]

Answer:

Oxidation number:

3*1+ oxidation number of J+2*-2= -1

Oxidation number of J = 0

5 0

2 years ago

What type of reaction is FeS + 2HCl àFeCl2 + H2S?

ollegr [7]

Answer: 3 4 a noodles

Explanation: just is $\lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \sqrt{x} x^{2} x^{2}$

7 0

2 years ago