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3 years ago

Please help me Solve this I DONT WANT THE ANSWER Just give me a good explanation...

Chemistry

1 answer:

timurjin [86]3 years ago

8 0

Answer:

Explanation:

'Ordered Arrangement' basically means that it is a solid at room temperature. Room temperature is approximately 15-20C so we are looking for melting and boiling points that are above room temperature so it hasn't/can't melt or boil at room temperature and would therefore be solid. Option C is the only one where both points are temperatures above room temperature therefore option C is the only one where the substance would be in an 'ordered arrangement' at room temperature.

Hope this helped!

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How many atoms of Kr (Krypton) are in a balloon that contains 2.00 mol of Kr? (4)

Svetllana [295]

Answer:

$atoms= 1.204x10^{24}atoms$

Explanation:

Hello!

In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:

$atoms=2.00mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms= 1.204x10^{24}atoms$

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2 years ago

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inna [77]

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$