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3 years ago

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White light shines on a piece of paper whose atoms are capable of absorbing orange, indigo, violet, yellow, red, blue, and green

. What color is the paper?

1 answer:

hram777 [196]3 years ago

5 0

Answer:

I think the color of the paper is Black.Am I right?

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Which class of organic compound is least likely to be used as an organic solvent? aldehyde ketone alkyl halide ether

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Aldehyde

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4 years ago

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Calculate how many grams of iron can be made from 16.5 grams of fe2o3 and 10 grams of h2 by fe2o2 3h2 to 2fe 3h2o find the limit

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3 years ago

The bonds of four salts (MgBr2, NaCl, MgCl2, and NaBr) are evaluated for ionic character based on electronegativity differences

sammy [17]

Answer:

MgBr2 < MgCl2 < NaBr < NaCl

Explanation:

In order to determinate ionic character, we have to subtract the electronegativity from the elements involved. This is known as Pauling rule.

MgBr2 > ΔΕΝ = 1.8

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3 years ago

At room temperature and pressure, 0.4 g of neon will occupy what volume<br> of gas?

LekaFEV [45]

Answer:

"One mole of any gas has a volume of 24 dm3 or 24,000 cm3 at rtp (room temperature and pressure). This volume is called the molar volume of a gas."

Explanation:

Divide the mass of neon.

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3 years ago

Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2

Elena-2011 [213]

Answer:

Approximately $0.224\;\rm L$ , assuming that this reaction took place under standard temperature and pressure, and that $\rm CO_2$ behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that $\rm CaCO_3$ is the limiting reactant. How many moles of $\rm CO_2$ would be produced?

Look up the relative atomic mass of $\rm Ca$ , $\rm C$ , and $\rm O$ on a modern periodic table:

$\rm Ca$ : $40.078$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm CaCO_3$ :

$\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $1\; \rm g$ of $\rm CaCO_3$ using its formula mass:

$\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}$ .

In the balanced chemical equation, the ratio between the coefficient of $\rm CaCO_3$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1$ .

In other words, for each mole of $\rm CaCO_3$ formula units consumed, one mole of $\rm CO_2$ would be produced.

If $\rm CaCO_3$ is indeed the limiting reactant, all that approximately $1.00\times 10^{-2}\; \rm mol$ of $\rm CaCO_3\!$ formula would be consumed. That would produce approximately $1.00\times 10^{-2}\; \rm mol\!$ of $\rm CO_2$ .

On the other hand, assume that $\rm HCl$ is the limiting reactant.

Convert the volume of $\rm HCl$ to $\rm dm^{3}$ (so as to match the unit of concentration.)

$\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $5.00\times 10^{-2}\; \rm dm^{3}$ of this $\rm 0.05\; \rm mol \cdot dm^{-3}$

$\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}$ .

Notice that in the balanced chemical reaction, the ratio between the coefficient of $\rm HCl$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}$ .

In other words, each mole of $\rm HCl$ molecules consumed would produce only $0.5\;\rm mol$ of $\rm CO_2$ molecules.

Therefore, if $\rm HCl$ is the limiting reactant, that $2.50 \times 10^{-3}\; \rm mol$ of $\rm HCl\!$ molecules would produce only one-half as many (that is, $1.25\times 10^{-3}\; \rm mol$ ) of $\rm CO_2$ molecules.

If $\rm CaCO_3$ is the limiting reactant, $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. However, if $\rm HCl$ is the limiting reactant, $1.25\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be produced.

In reality, no more than $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. The reason is that all $\rm CaCO_3$ would have been consumed before $\rm HCl$ was.

After finding the limiting reactant, approximate the volume of the $\rm CO_2\!$ produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately $22.4\; \rm L$ .

If $\rm CO_2$ behaves like an ideal gas, the volume of that $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be approximately $\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L$ .

3 0

3 years ago