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garik1379 [7]
3 years ago
6

Balance this equation: Ca(OH)2 + HCI → CaCl2 + H20

Chemistry
1 answer:
Verizon [17]3 years ago
5 0

Answer:

Double Displacement (Acid-Base)

Explanation:

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One of the products of photosynthesis is<br> carbon dioxide.<br> fire.<br> oxygen gas.<br> mass.
kondor19780726 [428]
One of the products of photosynthesis is carbon dioxide
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3 years ago
What mass of HgO is required to produce 0.692 mol of O2?<br><br>2HgO(s) -&gt; 2Hg(l) + O2(g)
Vika [28.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO  → 2 Hg + O_{2}

where,

HgO represents mercuric oxide

Hg represents mercury

O_{2} represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

       2×216 grams of HgO   →  1 mole of oxygen molecule

             ?                              →  0.692 moles of oxygen molecule

       

          = \frac{2*216*0.692}{1}

      = 298.944 grams of HgO

<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

<u />

7 0
3 years ago
Read 2 more answers
What is the molarity of 5.63 moles of lithium in 3.25 liters of solution
luda_lava [24]

Answer:

1.73 Molar

Explanation:

The formula is Molarity=moles of solute/liters of solution, which can be written in whatever way you prefer, and examples include: M=N/V or M=mol/L.

M=N/V

M= \frac{5.63}{3.25}

Divide 5.63 by 3.25. When you calculate this, you get 1.73, therefore your answer is 1.73 molar.

5 0
3 years ago
The diagram shows the electron arrangement in a molecule of ammonia, showing only outer
rusak2 [61]

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

3 0
3 years ago
I need help with this anyone?
valkas [14]
Just look it up on goog^le or a chart
5 0
3 years ago
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