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I am Lyosha [343]
4 years ago
14

During science class, while studying mixtures, you mix together table salt and sand. Your teacher challenges you to separate the

salt from the sand. What property could you use and how would you use it?
Chemistry
1 answer:
olga2289 [7]4 years ago
7 0
The property to be used is sieving. you get a bowl and a sieve when you pour the sand and salt,the salt being the small particle sieve to the bowl while the sand remain on the sieve as residue
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In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}

10=\frac{4}{3}*pi*{r_{L}}^{3}

{r_{L}}^{3}=\frac{7.5}{pi}

r_{L}=1.337cm

Step 2: Determine surface area of large sphere

A_{L}=4*pi*{r_{L}}^{2}

A_{L}=4*pi*1.337^{2}

A_{L}=22.463 cm^{2}

Step 3: Determine radius of small sphere

V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}

1.25=\frac{4}{3}*pi*{r_{s}}^{3}

{r_{s}}^{3}=\frac{0.9375}{pi}

r_{s}=0.668cm

Step 4: Determine surface area of small sphere

A_{s}=4*pi*{r_{s}}^{2}

A_{s}=4*pi*0.668^{2}

A_{s}=5.607 cm^{2}

Step 5: Determine total surface area of 8 small spheres

A_{S}=8*A_{s}

A_{S}=8*5.607

A_{S}=44.856 cm^{2}

A_{L}=22.463 cm^{2} - Surface area of 1 large sphere

A_{S}=44.856 cm^{2} - Surface area of 8 small spheres

Options:

  1. The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
  2. The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
  3. The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
  4. The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
  5. The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
  6. The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm
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Examples of a pure substances
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Answer:

Explanation:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.

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¿Cuántos electrones conforman un enlace doble?
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Answer:

Cada átomo aporta dos electrones al enlace, es decir, se comparten dos pares de electrones entre dos átomos. Un ejemplo es la molécula de Oxígeno (O2).

Explanation:

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What is the frequency of an electromagnetic wave that has a wavelength of
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➡ ANSWER

☑ <em><u>C</u></em><em><u>.</u></em><em><u> </u></em><em><u>3.5 105</u></em><em><u> </u></em><em><u>Hz</u></em>

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