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3 years ago

Is a leaf a solid,liquid or gas?

Chemistry

2 answers:

Virty [35]3 years ago

6 0

Answer:solid

Explanation:

Darina [25.2K]3 years ago

5 0

Answer:

It is a solid because leafs aren't made of water or gas.

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

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

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

The solubility of glucose at 30°C is

weqwewe [10]

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL \times \frac{0.996g}{1mL} =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

6 0

3 years ago

The rate constant for the decomposition reaction of H2O2 is 3.66 × 10−3 s−1 at a particular temperature. What is the concentrati

Pepsi [2]

Answer: 3.72 M

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.66\times 10^{-3}s^{-1}$

t = age of sample = 15.0 minutes

a = let initial amount of the reactant = 10.0 M

a - x = amount left after decay process = ?

$15.0\times 60s=\frac{2.303}{3.66\times 10^{-3}}\log\frac{10.0}{(a-x)}$

$\log\frac{100}{(a-x)}=1.43$

$\frac{100}{(a-x)}=26.9$

$(a-x)=3.72M$

The concentration of $H_2O_2$ in a solution after 15.0 minutes have passed is 3.72 M

7 0

3 years ago

Calculate the theoretical atom economy for each reaction.

Arisa [49]

100% find the gfm of both sides then divide

4 0

3 years ago

What is the concentration created when 380.0g of Pb(NO3)2 is added to 330.0 mL of solution?

Juliette [100K]

The concentration of lead nitrate is 3.48 M.

Explanation:

The molarity can be found by dividing moles of sucrose by its volume in litres. We can find the number of moles of sucrose by dividing the given mass by its molar mass. Now we can find the moles as,

Here mass of Pb(NO₃)₂ is 380 g

Molar mass of Pb(NO₃)₂ is 331.2 g/mol

Number of moles = $$\frac{given mass}{molar mass}$

= $$ \frac{380 g }{331.2 g/mol}$

= 1.15 moles

Volume in Litres = 330 ml = 0.33 L

Molarity = $$\frac{Moles}{Volume (L)}$

= 3.48 mol/L or 3.48 M

So the concentration of lead nitrate is 3.48 M.

7 0

3 years ago

Is a leaf a solid,liquid or gas?​

Is a leaf a solid,liquid or gas?