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3 years ago

Help mee please (I will give the person who answers this a brainlist)

Chemistry

2 answers:

Leokris [45]3 years ago

7 0

2 5 3 4 1 is the answers

Angelina_Jolie [31]3 years ago

5 0

Displacement is changing location
position is location
motion is your total motion no matter what direction your moving
reference point is how far you are from where you started
distance is using a object to compare your location

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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

Read 2 more answers

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

3. Give three examples of a pure substances

Nadya [2.5K]

Answer:

There will be weight, there will be volume, there will be height

Explanation:

5 0

3 years ago

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“Some types of vaccines need only one shot to protect you for the rest of your life. Other types, such as the one for seasonal f

IrinaK [193]

Yearly, is the answer.

8 0

3 years ago

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Can someone please help me with this?

BabaBlast [244]

Answer:

1. Main sequence stars have different masses. The common characteristic they have is their source of energy. They burn fuel in their core through the process of fusing hydrogen atoms into helium.

2. Supergiants are among the most massive and most luminous stars. Supergiant stars occupy the top region of the Hertzsprung–Russell diagram with absolute visual magnitudes between about −3 and −8. The temperature range of supergiant stars spans from about 3,400 K to over 20,000 K.

3. Supergiants develop when massive main-sequence stars run out of hydrogen in their cores.

4. a supernova occur When the pressure drops low enough in a massive star, gravity suddenly takes over and the star collapses in just seconds. This collapse produces the explosion.

5. when a star has reached the end of its life and explodes in a brilliant burst of light

Explanation:

4 0

3 years ago