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nlexa [21]
3 years ago
9

What impact would a sudden increase in the price of wood have on producers and consumer? How might each group respond to the cha

nge in price?
Chemistry
2 answers:
Lisa [10]3 years ago
6 0
The law of supply states, things remaining equal, an increase in price results in an increase in quantity supplied. The direct relation which is between the price and quantity. 

Therefore, when the price of wood increases, then the producers increases the quantity which is produced as they are willing to sell more at a higher price. Each group might respond to the change in price in that, the law of demand states that all things are equal.
A decrease in a quantity leads to an increase in price which is demanded. The existence in an inverse relation which is between price and quantity.
When the demand for good increases the consumer demand decreases.
Ket [755]3 years ago
4 0
The impact would make producers pay more increasing the pay of the workers and uping the price of the product. It would also slow the buying of wooden projects because of the Increase of price.
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A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL
AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL  of water. (Assume a density of 1.00 g/mL  for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

<u>Answer:</u> The molecular formula for the given organic compound is C_6H_{14}O_2

<u>Explanation:</u>

  • To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}

  • To calculate the depression in freezing point, we use the equation:

\Delta T_f=i\times K_f\times m

Or,

\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal boiling point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute = 10.20 g

M_{solute} = Molar mass of solute = ?

W_{solvent} = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g

<u>Calculating the molecular formula:</u>

We are given:

Percentage of C = 60.98 %

Percentage of H = 11.94 %

Percentage of O = (100 - 60.98 - 11.94) % = 27.08 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 60.98 g

Mass of H = 11.94 g

Mass of O = 27.08 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{60.98g}{12g/mole}=5.082moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.94g}{1g/mole}=11.94moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.08g}{16g/mole}=1.69moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.69 moles.

For Carbon = \frac{5.082}{1.69}=3

For Hydrogen = \frac{11.94}{1.69}=7.06\approx 7

For Oxygen = \frac{1.69}{1.69}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 7 : 1

The empirical formula for the given compound is C_3H_7O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 118.2 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

n=\frac{118.2g/mol}{59g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2

Hence, the molecular formula for the given organic compound is C_6H_{14}O_2

8 0
3 years ago
How many grams are there in a 14.5 moles lithium permanganate
frutty [35]

There are 1825.6 g in a 14.5 moles Lithium permanganate

<h3> Further explanation</h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

Moles can also be determined from the amount of substance mass and its molar mass :

\tt mol=\dfrac{mass}{MW}

moles of Lithium permanganate = 14.5

Lithium permanganate (LiMnO4) MW=125.9 g/mol, so mass :

\tt mass=mol\times MW\\\\mass=14.5\times 125.9~g/mol\\\\mass=1825.6~g

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Crest, trough, amplitude and wavelength 
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What would be the best source of data for you current diet?
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A diet rich in vegetables and fruits can lower blood pressure such as eating healthy
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Ion, any atom or group of atoms that bears one or more positive or negative electrical charges. Positively charged ions are called cations; negatively charged ions, anions.
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