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3 years ago

What are the best note taking styles for high school?

Chemistry

2 answers:

photoshop1234 [79]3 years ago

5 0

Answer:

cornell noted

Explanation:

used in middle school and can be helpful for subjects like ela history and science

marusya05 [52]3 years ago

4 0

Answer:

Cornell notes

(picture attached)

Explanation:

out of the 5 methods the most popular and efficient is Cornell notes :)

↓methods↓

1)The Cornell Method.

2)The Outlining Method.

3)The Mapping Method.

4)The Charting Method.

5)The Sentence Method

×║hope this helps║×

You might be interested in

50.0g of N2O4 is introduced into an evacuated 2.00L vessel and allowed to come to equilibrium with its decomposition product,N2O

Lady_Fox [76]

The mass of N2O4 in the final equilibrium mixture is 39.45 grams.

The decomposition reaction of N2O4 to 2NO2 can be expressed as:

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

From the parameters given:

The mass of N2O4 = 50.0 g
The molar mass of N2O4 = 92.011 g/mol

The number of mole of N2O4 can be determined as:

$\mathbf{Moles \ of \ N_2O_4 = \dfrac{50.0 g}{92.011 g/mol}}$

$\mathbf{Moles \ of \ N_2O_4 = 0.5434 \ moles}$

The volume of the vessel in which N2O4 was evacuated is = 2.0 L

From stochiometry, the concentration of $\mathbf{[N2O4] = \dfrac{mole \ of \ N_2O_4}{volume \ of \ N_2O_4}}$

$\mathbf{[N2O4] = \dfrac{0.5434}{2}}$

$\mathbf{[N2O4] = 0.2717 \ M}$

The I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.2717 0

Change -x +2x

Equilibrium (0.2717 - x) 2x

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(2x)^2}{(0.2717-x)}}$

Recall that; Kc = 0.133

∴

$\mathbf{0.133 = \dfrac{4x^2}{(0.2717-x)}}$

0.0361 - 0.133x = 4x²

4x² + 0.133x - 0.0361 = 0

By solving the above quadratic equation, we have;

x = 0.07978

The Concentration of [NO2] = 2x

[NO2] = 2 (0.07978)
[NO2] = 0.15956 M

The Concentration of [N2O4] = 0.2717 - x

[N2O4] = 0.2717 - 0.07978
[NO2] = 0.19192 M

Again, from the decomposition reaction, we can assert that;

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.

However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;

$\mathbf{= \dfrac{5.0 \ g}{46 \ g/mol}} \\ \\ \\ \mathbf{= 0.10869\ moles}$

The Molarity of NO2 injected now becomes:

$\mathbf{= \dfrac{00.10869 }{2} } \\ \\ \\ \mathbf{= 0.05434 \ M }$

So, the new moles of [NO2] becomes = 0.15956 + 0.05434

= 0.2139 M

The new I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.19192 0.2139 M

Change +x -2x

Equilibrium (0.19192 + x) (0.2139 -2x)

NOTE: The injection of NO2 makes the reaction proceed in the backward direction.

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

Recall that; Kc = 0.133

$\mathbf{0.133= \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

By solving for x;

x = 0.2246 or x = 0.0225

We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:

x = 0.0225

Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M

= 0.21442 M

The final number of moles of N2O4 = Molarity(concentration) × volume

The final number of moles of N2O4 = (0.21442 × 2) moles

The final number of moles of N2O4 = 0.42884 moles

The mass of N2O4 in the final equilibrium mixture is:

= final number of moles × molar mass of N2O4

= 0.42884 moles × 92 g/mol

= 39.45 grams

Learn more about the decomposition of N2O4 here:

brainly.com/question/25025725

6 0

3 years ago

Can someone help pls! I’m very confused on this!!

Klio2033 [76]

1. 160,000
2. 160,00
You just have to multiply the volume and the buffer.

6 0

3 years ago

A boy found a solid metal box in his backyard. The box had been buried for so long, it was difficult to determine from what the

Alex777 [14]

Answer:

Box is made up of <em>copper</em>, because density is <em>8.96 g/cm³.</em>

Explanation:

Given data:

Volume of box = 17.63 cm³

Mass of box = 158 g

Which metal box is this = ?

Solution:

First we will calculate the density of box then we will compare it with the density value of given metals.

d = m/v

d = 158 g/ 17.63 cm³

d = 8.96 g/cm³

The calculated density is similar to the given density value of copper thus box is made up of copper.

4 0

4 years ago

The average propane cylinder for a residential grill holds approximately 18 kg of propane. how much energy (in kj) is released b

Angelina_Jolie [31]

Let's begin with the basic values that will be used in the solution.

The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.

Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(C3H8)= 3*12+ 8*1= 44 g

Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (ΔHc°).

ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).

Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) ΔHc° = -2202 kj/mol

Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole

1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj

The answer is 1000909 kj.

6 0

4 years ago

Which disease is caused by a living pathogen? a. common colds b. smallpox c. cholera d. influenza

liraira [26]

Answer:

Infectious diseases are caused by pathogens, which include bacteria, fungi, protozoa, worms, viruses, and even infectious proteins called prions

7 0

3 years ago