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2 years ago

What are two reasons that ICl has a low melting point?

Chemistry

1 answer:

vovangra [49]2 years ago

3 0

Answer:

1. It is a covalent bond and all covalent bonds have low melting points

Explanation:

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What mineral is most likely used to make the MP3 player?

Vlad [161]

I am going to gamble with Quartz as the material is just the same... As zinc is used to make batteries, It has to be quartz hands down.

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3 years ago

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Chemical Equations

vagabundo [1.1K]

Answer:

0.712 moles of NO₂ are formed.

Explanation:

First, we need to write the balanced equation:

2 N₂O₅(g) ⇄ 4 NO₂(g) + O₂(g)

From the balanced equation, we can see the relationship between the moles of N₂O₅ and the moles of NO₂. Every 2 moles of N₂O₅ that react, 4 moles of NO₂ are formed. Let us apply this relationship to the information given by the problem (0.356 moles of N₂O₅):

$0.356molN_{2}O_{5}.\frac{4molNO_{2}}{2molN_{2}O_{5}} =0.712molNO_{2}$

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3 years ago

Which two transitions can magma undergo?

UNO [17]

Crystallization above the ground and crystallization below the ground

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3 years ago

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

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3 years ago

3. How many molecules are in 0.500 moles of sulfur?

satela [25.4K]

Explanation:

No of molecules=0.500×6.023×10²³=3.011×10²³ molecules

7 0

3 years ago