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Citrus2011 [14]

3 years ago

9

hen car tires are on hot pavement for too long, the pressure inside will _____ and the volume within the tire will _

______.

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Your answer your answer correct

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Who is telling what happens in Home Début?

nekit [7.7K]

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3 years ago

Plzz!!! I need help now<br><br><br> this is science

Neko [114]

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3 years ago

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Rina8888 [55]

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Temporary hardness is a type of water hardness caused by the presence of dissolved bicarbonate minerals (calcium bicarbonate and magnesium bicarbonate). ... However, unlike the permanent hardness caused by sulfate and chloride compounds, this "temporary" hardness can be reduced by boiling the water.

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3 years ago

a certain piece of copper wire is determined to have a mass of 2.0 0g per meter how many centimeters of the wire would be needed

harina [27]

The given question tells you that a certain piece of wire has a mass = 2.0 g per meter

That means that if you consider a piece of wire that is 1 m in length, its mass will be equal to 2.00 g .

According to question,

Now, you know that

1 m = 100 cm

Mass of 2.00 g of copper will correspond to a wire that is a 100 cm long.

This implies that 0.28 gof copper will correspond to a wire that is

0.28 g * 100 cm / 2.0 g = 14 cm long

Hence, 14 cm of the wire would be needed to provide 0.28 g of copper.

Properties of Copper :

High conductivity and ductility.

Non magnetic.

To know more about Copper here:

brainly.com/question/19761029?referrer=searchResults

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3 0

1 year ago

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