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4 years ago

What is wind average

Chemistry

2 answers:

Reika [66]4 years ago

7 0

Wind average is between 6and 12 miles per hour and 19 kilometers per hour

-Dominant- [34]4 years ago

5 0

Wind average is the daily average wind speed.

In the U.S., daily wind speeds typically average between 6 and 12 miles per hour (10 and 19 kilometers per hour) throughout the course of a year. These averages vary widely by geographic location.

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Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below. 2HgO Rig

BigorU [14]

Taking into account the reaction stoichiometry, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.

In first place, the balanced reaction is:

2 HgO → 2 Hg + O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

HgO: 2 moles
Hg: 2 moles
O₂: 1 moles

The molar mass of the compounds is:

HgO: 216.59 g/mole
Hg: 200.59 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

HgO: 2 moles× 216.59 g/mole= 433.18 grams
Hg: 2 moles× 200.59 g/mole= 401.18 grams
O₂: 1 mole× 32 g/mole= 32 grams

Then the following rule of three can be applied: if by reaction stoichiometry 32 grams of O₂ are produced by 2 moles of HgO, 250 grams of O₂ are produced from how many moles of HgO?

$moles of HgO=\frac{250 grams of O_{2} x2 moles of HgO}{32grams of O_{2}}$

moles of HgO= 15.625 moles≅ 15.63 moles

Finally, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.

Learn more about reaction stoichiometry:

5 0

3 years ago

How many atoms are present in 179.0 g of iridium? include units please:)!

denpristay [2]

Answer:

179.0 g of iridium (1 mol / 192.217 g) ( 6.022 x 10^23 atoms / 1 mol ) = 5.61 x 10^23 atoms of iridium

Explanation:

5 0

4 years ago

Read 2 more answers

52. Which is true about all the different types of matter?

Svetllana [295]

Types of matter are just different ways molecules behave at certain temperatures. There's no specific atoms or molecules that make solids, liquids, or gases. Ice, water, and water vapor are all H2O, for instance. The only correct answer would be that it is made of the same basic particles, but then again, so is everything.

7 0

4 years ago

Need help balancing these

kari74 [83]

The answer would be 1,3,1,3

3 0

3 years ago

Read 2 more answers

what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield

earnstyle [38]

Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

$\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles$

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

$TiS+H_2O\rightarrow TiO+H_2S$

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

$\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}$

$\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g$

To calculate the percentage yield of titanium (II) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

$\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%$

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

5 0

3 years ago