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2 years ago

7

Sarah heated some orange coloured crystals in a test tube. The crystals turned green and fizzed. After the fizzing stopped, the

test tube was allowed to cool. A green powder was left in the tube. Determine whether a chemical or a physical reaction has occurred. Identify two observations that support your decision.

1 answer:

Anni [7]2 years ago

6 0

A chemical change has occurred.
This is indicated by a colour change occurring (from orange to green) as well as a gas being given off (seen as the fizzing). If the fizzing made a small amount of noise, it also indicates that sound energy is given off, which is another indication that a chemical reaction has occurred.

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A substance joined by this type of bond will conduct electricity when in?

Luba_88 [7]

Answer:

C.) If the solid is conductive, the bonds are metallic.

1) Substances with ionic bond conduct an electric current in liquid, but not in solid state, because in liquids ions are mobile, contrary to solids where ions are fixed.

2) Substances with covalent bond not conduct an electric current in liquid and solid state, because they not have free ions or electrons.

3) Substances with metallic bond conduct an electric current in liquid and solid state, because they have mobile electrons. Most metals have strong metallic bond, because strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.

Explanation:

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2 years ago

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

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

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

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2 years ago

What is the hydrophobic effect? what types of atoms are most likely to be found in portions of molecules or entire molecules tha

HACTEHA [7]

Hydrophobic in simple terms is the "fear" of water.

In chemistry, Ionic salts are technically hydrophilic as they will dissolve in water.
most fats and oils or triesters won't why?

salts are made from ions and their structure contain polar bonds Na+ and Cl-
Fats and oils tend to not have these.

Fats and oils contain many carbon covalent bonds which are non-polar which tend to make the molecule non-polar as a whole, therefore, it can't dissolve in water and is said to be hydrophobic.

Hope that heps :)
NB included a diagram to help.

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3 years ago

Helga [31]

Metallic I’m pretty sure. :)

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3 years ago

Is the process by which certain waste materials, such as plastic, glass, aluminum, and paper, are prepared for reuse.

andriy [413]

the process is called recycling.

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3 years ago

Read 2 more answers