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mars1129 [50]
3 years ago
5

In an experiment, zinc chlorate decomposed according to the following chemical equation.

Chemistry
2 answers:
-BARSIC- [3]3 years ago
8 0

Answer:

The last one is the correct answer!

Explanation:

I just took the test ;)

mote1985 [20]3 years ago
6 0

Answer : The correct expression is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Explanation :

First we have to calculate the moles of Zn(ClO_3)_2

\text{ Moles of }Zn(ClO_3)_2=\frac{\text{ Mass of }Zn(ClO_3)_2}{\text{ Molar mass of }Zn(ClO_3)_2}=\frac{150g}{232.29g/mole}=\frac{150}{232.29}mole

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

Zn(ClO_3)_2\rightarrow ZnCl_2+3O_2

From the balanced chemical reaction we conclude that,

As, 1 mole of Zn(ClO_3)_2 react to give 3 mole of O_2

So, \frac{150}{232.29} moles of Zn(ClO_3)_2 react to give \frac{150}{232.29}\times \frac{3}{1} moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(\frac{150}{232.29}\times \frac{3}{1})mole\times (31.998g/mole)=\frac{150\times 3\times 31.998}{232.29\times 1}g

Therefore, the correct expression for the mass of oxygen gas formed is, (150 × 3 × 31.998) ÷ (232.29 × 1) grams

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How many half-lives would something have gone through if you had 75% daughter product and 25% parent?
levacccp [35]

Answer:

Two Half-lives

Explanation:

Let number of Parent nuclei Initially present be X,

Then, finally \frac{X}{4} Parent nuclei Will remain with \frac{3X}{4} daughter nuclei.

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After another half life , Parent nuclei will become half of \frac{X}{2}

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30.0 ml of an hf solution were titrated with 22.15 ml of a 0.122 m koh solution to reach the equivalence point. what is the mola
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Answer is: molarity of hydrofluoric solution is 0.09 M.

Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
V(HF) = 30.0 mL.
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V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
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