Which three elements affect air pressure?

Write an equation that shows the formation of a lithium ion from a neutral lithium atom

AleksAgata [21]

Li + e- ===> Li+
it looses its valence electron to form a uni-positive ion

8 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

3 years ago

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What is the mass of 200 kg object

-Dominant- [34]

M = 200 Kg

F = 2600 N

A = ?

Solution:

F = MA

A = F/M

A= 13 m/s^2

Im not sure but i tired

4 0

3 years ago

How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric aci

horsena [70]

Answer:

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

15 g / 36.45 g/m = 0.411 of acid

1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g

5 0

3 years ago

Hii pls help me to balance the equation thanksss

Vlada [557]

Answer:

3Fe(s)+2O2(g)---->Fe3O4

this way you will have 3irons on both sides and 4 oxygens.

I hope this helps

3 0

3 years ago

Read 2 more answers