Which of the following statements is a testable hypothesis for an experiment

Chemistry

2 answers:

Arlecino [84]3 years ago

6 0

Answer:

Explanation:

There are many different testable hypothesis and experiments that can be made in order to prove this. Since there are no available statements let me provide one.

Testing and Comparing the excrement of wild and tamed horses.

This is because the excrement of tame horses will show higher levels of man-made chemicals and additives due to the different man-made horse food that the owners are giving them. Wild horses on the other hand will not since they nourish themselves on naturally grown grass.

valentinak56 [21]3 years ago

3 0

The teeth of wild horses wear down faster than the teeth of a tame horse.

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For the reaction 3ZnCO3 + 2H3PO4 —> 3H2O + 3CO2, how many miles of zinc carbonate are required to produce 3.0 mil of Zn3(PO4)

drek231 [11]

Answer:

$\boxed{ \text{9.0 mol of ZnCO$_{3}$}}$

Explanation:

(a) Balanced equation

3ZnCO₃ + 2H₃PO₄ ⟶ Zn₃(PO₄)₂ + 3H₂O + 3CO₂

(b). Calculation

You want to convert moles of Zn₃(PO₄)₂ to moles of ZnCO₃

The molar ratio is 1 mol Zn₃(PO₄)₂:3 mol ZnCO₃

$\text{Moles of ZnCO$_{3}$} =\text{3.0 mol Zn$_{3}$(PO$_{4})_{2}$ } \times \dfrac{\text{3.0 mol ZnCO$_{3}$}}{\text{1 mol Zn$_{3}$(PO$_{4}$)$_{2}$}} \\\\\\= \text{9.0 mol ZnCO$_{3}$}$

You need $\boxed{ \textbf{9.0 mol of ZnCO$_{3}$}}$ to form 3.0 mol of Zn₃(PO₄)₂.

4 0

3 years ago

the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin

Vanyuwa [196]

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = $\frac{30 * 3}{2}\\$ = 45 molecules of Cl₂.