All categories

3 years ago

Describe how the surface area affect the rate of a reaction

1 answer:

7 0

When we increase the surface area of an object, more atoms are exposed. Since more atoms are exposed, the atoms can react faster, and this is why the rate of a reaction increases when the surface area increases.

For example, lets say we want to heat a potato. If we just put the whole potato in the microwave, it will take a long time for the potato to get thoroughly heated. However, if we chop the potato into smaller pieces, we will observe that it gets heated much faster. This is because we increased the surface area of the potato, which resulted in more potato atoms to be exposed to the heat, and caused the reaction to be faster.

You might be interested in

Compare the mass of the Mg ribbon with the mass of the magnesium oxide. Notice that the the mass of the magnesium oxide is great

Artist 52 [7]

Answer:

Magnesium oxide is a binary compound of magnesium and oxygen while magnesium ribbon consists only of magnesium atoms.

Explanation:

The burning of magnesium in oxygen is a chemical change. It produces magnesium oxide having greater mass than magnesium ribbon. The greater mass results from the fact that the chemical reaction has added another element to the sample- oxygen. The mass of magnesium ribbon is the mass of magnesium atoms alone but in magnesium oxide, we consider the masses of magnesium and oxygen atoms making magnesium oxide heavier than magnesium ribbon.

4 0

3 years ago

Help(Must be done by 2/22/2018)

alekssr [168]

These are 6 questions and 6 answers.

Question 1:

Answer: 33.7 atm

Explanation:

1) Data:

p=?
m = 1360.0 g N2O
V = 25.0 liter
T = 59.0°C

2) Formulas:

Ideal gas law: p V = n R T
n = mass in grams / molar mass

3) Solution

n = mass of N2O in grams / molar mass of N2O

molar mass of N2O = 2 * 14 g/mol + 16 g/mol = 44 g/mol

n = 1360.0 g / 44 g/mol = 30.9 mol

T = 59.0 + 273.15 K = 332.15 K

R = 0.0821 atm*liter / K*mol

=> p = nRT / V = 30.9 mol * 0.0821 [atm*liter / K * mol] * 332.15K / 25.0 liter = 33.7 atm

Answer: 33.7 atm

Question 2:

Answer: 204.5 liter

Explanaton:

1) Data:

m = 11.7 g of He
V = ?
p = 0.262 atm
T = - 50.0 °C

2) Formulas:

pV = nRT

n = mass in grams / atomic mass

3) Solution:

atomic mass of He = 4.00 g/mol

n = 11.7 g / 4.00 g/mol = 2.925 mol

T = - 50.0 + 273.15 K = 223.15 K

pV = nRT => V = nRT / p

V = 2.925 mol * 0.0821 [* liter / K*mol] *223.15K / 0.262 atm = 204.5 liter

Answer: 204.5 liter

Question 3.

Answer: 97.8 mol

Explanation:

1) Data:

Ethane
T = 15.0 °C
p = 100.0 kPa
V = 245.0 ml
n = ?

2) Formula

pV = nRT

3) Solution

pV = nRT => n = RT / pV

T = 15.0 + 273.15K = 288.15K

R = 8.314 liter * kPa / (mol*K)

n = 8.314 liter * kPa / (mol*K) * 288.15K / [100.0 kPa * 0.245 liter] = 97.8 mol

Answer: 97.8 mol

Question 4:

Answer: 113.67 K = - 159.48 °C

Explanation:

1) Data:

V = 629 ml of O2
p = 0.500 atm
n = 0.0337 moles
T = ?

2) Formula:

pV = nRT

3) Solution:

pV = nRT => T = pV / (nR)

T = 0.500 atm * 0.629 liter / (0.0337 mol * 0.0821 atm*liter/K*mol ) = 113.67 K

°C = T - 273.15 = - 159.48 °C

Question 5.

Answer: 5.61 g

Explanation:

1) Data:

V = 3.75 liter of NO
T = 19.0 °C
p = 1.10 atm
m = ?

2) Formulas

pV = nRT

mass = number of moles * molar mass

3) Solution:

pV = nRT => n = pV / (RT)

T = 19.0 + 273.15 K = 292.15 K

n = 1.10 atm * 3.75 liter / [ (0.0821 atm*liter / K*mol) * 292.15 K ] = 0.17 mol

molar mass of NO = 17.0 g/mol + 16.0 g/mol = 33.0 g/mol

mass = 0.17 mol * 33.0 g/mol = 5.61 g

Question 6:

Answer: 22.4 liter

Explanation:

1) Data:

STP
n = 1.00 mol
V = ?

Solution:

1) It is a notable result that 1 mol of gas at STP occupies a volume of 22.4 liter, so that is the answer.

2) You can calculate that from the formula pV = nRT

3) STP stands for stantard pressure and temperature. That is p = 1 atm and T = 0°C = 273.15 K

4) Clear V from the formula:

V = nRT / p = 1.00 mol * 0.0821 atm*liter / (K*mol) * 273.15 K / 1.00 atm = 22.4 liter

7 0

3 years ago

4 kmol Methane (CH4) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both

Svetach [21]

Answer:

Heat transfer = 3564 Jolues

The same value

Explanation:

The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):

Qtransfer = - mol x ΔHºc Qtransfer

So look up in appropiate reference table ΔHºc and solve the problem:

ΔHºc = - 891 kJ/mol

Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J

if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.

7 0

3 years ago

In general, when the addition of an unsymmetrical electrophilic reagent to an unsymmetrical alkene forms the product predicted b

Tju [1.3M]

It is formed via the more/most stable carbocation.

What justification exists for Markovnikov's Rule?

A carbocation is created as a result of the protonation of the alkene by the protic acid. The carbocation that has the most alkyl substituents on its carbon holds the positive charge, making it the most stable carbocation. As a result, the addition of the halide to the carbon that has fewer hydrogen substituents makes up the majority of the product.

Markovnikov's rule

According to the Markovnikov rule, a proton is added to the carbon atom with the greatest number of hydrogen atoms connected in addition to processes involving alkenes or alkynes.

Anti-Markovnikov Rule

According to the Anti-Markovnikov Rule, in addition to reactions between alkenes or alkynes, The carbon atom with the fewest hydrogen atoms linked to it receives the proton.

Learn more about Carbocation here:-

brainly.com/question/14363695

#SPJ4

7 0

2 years ago

Denise owns a car that runs on a mixture of gasoline and ethanol. She can buy fuels that have 85% ethanol or 25% ethanol. How mu

Hunter-Best [27]

<span>85% ethanol | 25% ethanol | 50% ethanol
x | y | 20 gal

use x and y because you don;t know how much she needs.

0.85x | 0.25y | 20(0.5)

85% is 85/100 or 0.85, and you need that much of x, same goes for the 25% and 50% mixtures so now you can make up 2 equations
1) x + y = 20 2) 0.85x + 0.25y= 10 (you get 10 when you multiply 20 by 0.5) now you can solve for x or y using substitution.
first rewrite 1) in terms of x or y: x+ y= 20 ----> y= 20 - x now you can substitute 20- x for y in the second equation.. 0.85x + 0.25y= 10 0.85x + 0.25(20-x)= 10 distribute here..(0.25 * 20 and 0.25 * (-x) ) 0.85x + 5 - 0.25x = 10 combine like terms 0.6x +5 = 10 move the 5 over to the other side 0.6x= 10 -5 0.6x = 5 divide both sides by 0.6 x= 25/3 or 8.3 now you know the amount of x so you can substitue this back into the first equation to find y. 0.85x + 0.25y= 10 0.85(25/3) +0.25y= 10 85/12 + 0.25y= 10 0.25y = 10- 85/12 0.25y= 35/12 y= 35/3 or 11.6 you can check by putting these values into the euations: 1) x+ y= 20 25/3 + 35/3 =20 20= 20 good so far 2) 0.85x + 0.25y= 10 0.85(25/3) + 0.25(35/3)=10 10 = 10

so our values for x and y work
x= 25/3 and y= 35/3</span>

5 0

3 years ago