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motikmotik
2 years ago
9

Describe how you would find out whether copper(ii) oxide was a catalyst for the decomposition of hydrogen peroxide solution. You

need to show not only that it speeds the reaction up, but that it is chemically unchanged at the end.​
Chemistry
1 answer:
Alik [6]2 years ago
8 0

Answer:

ds

Explanation:

sds

You might be interested in
All + HgCl, --------> AICI,
Free_Kalibri [48]

Answer:

64.7g

Explanation:

The balanced chemical equation of this question is as follows;

AlI + HgCl2 → HgI + AlCl2

Based on the above equation, 1 mole of AlI (aluminum monoiodide) reacts to produce 1 mole of HgI (mercury iodide).

Using mole = mass/molar mass to convert mass of HgI to moles.

Molar mass of HgI = 200.59 + 127

= 327.59g/mol

Mole = 138/327.59

= 0.42mol

- If 1 mole of AlI (aluminum monoiodide) reacts to produce 1 mole of HgI (mercury iodide)

- Then 0.42 mol of HgI will be produced by 0.42mol of AlI.

Using mole = mass/molar mass

Mass = mole × molar mass

Molar mass of AlI = 27 + 127

= 154g/mol

Mass of AlI = 0.42 × 154

= 64.7g of AlI

5 0
3 years ago
Meat-eating animals are:
Rina8888 [55]

Answer:

<em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em>-</em><em>o</em><em>r</em><em>d</em><em>e</em><em>r</em><em> </em><em>c</em><em>o</em><em>n</em><em>s</em><em>u</em><em>m</em><em>e</em><em>r</em><em>s</em><em>.</em>

Explanation:

<em>c</em><em>a</em><em>r</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>o</em><em>m</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em>a</em><em>r</em><em>y</em><em> </em><em>c</em><em>o</em><em>n</em><em>s</em><em>u</em><em>m</em><em>e</em><em>r</em><em>s</em><em>.</em><em> </em><em>m</em><em>a</em><em>n</em><em>y</em><em> </em><em>c</em><em>a</em><em>r</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em> </em><em>e</em><em>a</em><em>t</em><em> </em><em>h</em><em>e</em><em>r</em><em>b</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em>.</em><em> </em><em>s</em><em>o</em><em>m</em><em>e</em><em> </em><em>e</em><em>a</em><em>t</em><em> </em><em>o</em><em>m</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>s</em><em>o</em><em>m</em><em>e</em><em> </em><em>e</em><em>a</em><em>t</em><em> </em><em>o</em><em>t</em><em>h</em><em>e</em><em>r</em><em> </em><em>c</em><em>a</em><em>r</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em>.</em><em> </em><em>c</em><em>a</em><em>r</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em> </em><em>t</em><em>h</em><em>a</em><em>t</em><em> </em><em>c</em><em>o</em><em>n</em><em>s</em><em>u</em><em>m</em><em>e</em><em> </em><em>o</em><em>t</em><em>h</em><em>e</em><em>r</em><em> </em><em>c</em><em>a</em><em>r</em><em>n</em><em>i</em><em>v</em><em>o</em><em>r</em><em>e</em><em>s</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>c</em><em>a</em><em>l</em><em>l</em><em>e</em><em>d</em><em> </em><em>t</em><em>e</em><em>r</em><em>t</em><em>i</em><em>a</em><em>r</em><em>y</em><em> </em><em>c</em><em>o</em><em>n</em><em>s</em><em>u</em><em>m</em><em>e</em><em>r</em><em>s</em><em>.</em>

8 0
2 years ago
Read 2 more answers
Does the Caribbean have high or low viscosity lava?
butalik [34]
High now I need to write 20 characters
8 0
3 years ago
A chemist prepares a solution of sodium chloride(NaCl) by measuring out 25.4g of sodium chloride into a 100ml volumetric flask a
labwork [276]

Answer: 4 molL-1

Explanation:

Detailed solution is shown in the image attached. The number of moles of NaCl is first obtained. Since the molarity must be in units of molL-1, the volume is divided by 1000 and the formula stated in the solution is applied and the answer is given to one significant figure.

5 0
3 years ago
Read 2 more answers
A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to
Simora [160]

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

8 0
3 years ago
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