Hello, I need help with this question

Engineers increase output in crop production by combining the strengths of

katrin [286]

Answer:

Cross-pollination

Explanation:

This technique is used to create new varieties, usually with stronger characteristics than the two original varieties. When the pollen of one variety is placed to a different plant's stigma, the resulting seeds create a new variety. This cross-pollination is used in plants of the same species.

5 0

3 years ago

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

2 years ago

Read 2 more answers

What is the pH of a 2.10 x 10⁻⁶ M solution of HCl?

Delvig [45]

Answer:

pH = 5.7

Explanation:

pH = -log[H^+]

For HCl pH = -log[HCl] = - log [2.10 x 10⁻⁶ ] = 5.7

5 0

2 years ago

What is electron shielding

Artist 52 [7]

Shielding electrons are the electrons in the energy levels between the nucleus and the valence electrons. They are called "shielding" electrons because they "shield" the valence electrons from the force of attraction exerted by the positive charge in the nucleus. Hope this helps!!

8 0

3 years ago

Read 2 more answers

Name 3 objects found in the Milky Way galaxy.

dolphi86 [110]

Answer:

The Earth, Stars, and Mars I guess...

5 0

2 years ago