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schepotkina [342]
4 years ago
8

What are scientific models used for? Give an example of each of the following types of models: Idea, Physical, Compute

Chemistry
1 answer:
7nadin3 [17]4 years ago
7 0

Answer:

Scientific models are used to explain and predict the behaviour of real objects or systems and are used in a variety of scientific disciplines, ranging from physics and chemistry to ecology and the Earth sciences.

Explanation:

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A solution is made by dissolving 4.35 g of glucose (C6H1206) in 25.0 mL of water at 25 °C.Calculate the molality of glucose in t
Roman55 [17]

Answer:

The molality is unchanged (0.96 molal)

Explanation:

<u>Step 1: </u>Data given

mass of glucose = 4.35 grams

volume of water = 25.0 mL

Density of water = 1.00 g/mL

Molar mass of glucose = 180.156 g/mol

<u>Step 2:</u> Calculate number of moles

moles of glucose = mass of glucose / Molar mass of glucose

moles of glucose = 4.35 grams / 180.156 g/mol

moles of glucose = 0.024 moles

<u>Step 3:</u> Calculate mass of water

mass = density * volume

mass of water = 1.00 g/mL * 25.0 mL

mass of water = 25 g = 0.025 kg

<u>Step 4</u>: Calculate molality

molality = Number of moles / mass of water

molality = 0.024 moles / 0.025 kg

molality = <u>0.96 molal</u>

Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled.

This means double mass of water = 2*0.025 kg = 0.050 kg

Now molality is 0.024 moles / 0.050 kg = 0.48 molal

When the mass of solvent is doubled, the molality is halved from 0.96 molal to <u>0.48 molal</u>

You take this newsolution and add more solute, so that the original mass of the solute is doubled.

This means double mass of glucose = 2*4.35 g = 8.70 g

8.70 grams of glucose = 8.7 grams * 180.156 g/mol = 0.048 moles

molality = 0.048 moles / 0.050 kg = <u>0.96 molal</u>

The molality is unchanged

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4 years ago
What is the volume of 5.00 mole of an unknown gas at STP
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Answer:

What is the volume of gas at 2.00 atm and 200.0 K if its original volume was. 200.0 L at STP  How many moles of nitrogen gas will occupy a volume of 150 L at STP? n=pr 11.00atm) What is the mass of 5.00 L of NO2 gas at STP? PV = nRT.

Explanation:

5 0
3 years ago
A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo
Kruka [31]

<u>Answer:</u> The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

To calculate the molarity of solution, we use the equation:

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

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M

Hence, the molarity of sucrose solution is 0.146 M

  • <u>Calculating the molality of solution:</u>

To calculate the mass of solvent, we use the equation:

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

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g

To calculate the molality of solution, we use the equation:

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (sucrose) = 15 g

M_{solute} = Molar mass of solute (sucrose) = 342.3 g/mol

W_{solvent} = Mass of solvent = 339 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m

Hence, the molality of sucrose solution is 0.129 m

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4 years ago
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