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ira [324]
2 years ago
14

The empirical formula for a compound that is 1.2% h, 42.0%cl, and 56.8%o is

Chemistry
1 answer:
Charra [1.4K]2 years ago
7 0
Hope you understand how to work out those types of questions now xD ;)

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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
Which activity might help to increase the validity of this experiment?
MArishka [77]
A should be the answer because the more you test an experiment the more data you have to rely on changing the experiment would cause you to have different outcomes making the results different and unreliable so B, C, and D is not going to be the answer Hope this helps
5 0
2 years ago
At a given temperature, the elementary reaction A --->B in the forward direction is first order in A with a rate constant of
Pani-rosa [81]

Answer:

The value of the equilibrium constant for the reaction A ⇒ B is Kc = 1.72 × 10³.

The value of the equilibrium constant for the reaction B ⇒ A is K'c = 5.81 × 10⁻⁴.

Explanation:

For the reaction A ⇒ B, the equilibrium constant (Kc) is equal to the forward rate constant (kf) divided by the reverse rate constant (ki).

Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1}   }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}

If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.

K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3}  } =5.81 \times 10^{-4}

4 0
3 years ago
Before a child takes a bath, his mother mixes bubbles into bath water. Physical or Chemical Change?
NeX [460]

Answer:

chemical

Explanation:

6 0
3 years ago
Which of the following correctly describes the state of matter of one of the samples? D is a liquid because it retains its shape
Brut [27]
D is a solid because it retains its shape
7 0
2 years ago
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