A short true-breeding plant will always produce a short plant.<br> True<br> False

If you lose half of your mass, its gravity will.. causing the tides to ..

babymother [125]

Muchas gracias amigo

5 0

3 years ago

14. How many liters would 5 moles of oxygen gas take up?

andrew-mc [135]

Answer:

a 0.22L

Explanation:

8 0

3 years ago

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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:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will be slower - false, 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
The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

7 0

3 years ago

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How many grams of oxygen is needed to completely react with 9.30 moles of aluminum

Orlov [11]

Balanced chemical equation:

* moles of oxygen

4 Al + 3 O2 = 2 Al2O3

4 moles Al -------------- 3 moles O2
9.30 moles Al ---------- moles O2

moles O2 = 9.30 * 3 / 4

moles O2 = 27.9 / 4 => 6.975 moles of O2

Therefore:

Molar mass O2 = 31.9988 g/mol

n = m / mm

6.975 = m / 31.9988

m = 6.975 * 31.9988

m = 223.19 of O2

5 0

3 years ago

What change would occur in the kinetic energy of vapor if a container of vapor is placed at 0 Kelvin?

Vlada [557]

Partical will have no kinetic energy

5 0

4 years ago

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A short true-breeding plant will always produce a short plant. True False