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Elan Coil [88]
4 years ago
11

What is the percent by mass of oxygen in al2o3 ?

Chemistry
1 answer:
Snezhnost [94]4 years ago
4 0
mr oxygen= 16. 16 x 3= 48
mr Al2O3= 27 + 27 + 48= 102

(48/102) X 100%= 47.05% oxygen
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12 Describe In your own words, describe the glassblowing process.
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2 years ago
How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh
Sever21 [200]

Answer:

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

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Step 3:

Determination of the volume of the acid.

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Cross multiply to express in linear form

6 x Va = 2 x 39

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Va = (2 x 39)/6

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Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL

5 0
3 years ago
Predict whether ΔS° is greater than, less than, or approximately zero for each of the following reactions, and explain your choi
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Answer:

Explanation:

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And

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As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

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From the question ,

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8 0
3 years ago
What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according
ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

<u>Step 1: </u>The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

<u>Step 2: </u>Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

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<u>Step 3: </u>Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

<u>Step 4: </u>Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M  = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0
3 years ago
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