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STatiana [176]
3 years ago
10

When a jar of coffee is opened, people in all parts of the room soon notice the smell.

Chemistry
2 answers:
ser-zykov [4K]3 years ago
8 0

Answer:

Explanation:

because in air the particles are moving very fast and jar of coffee has hot particles which kinetic energy is also more so that's why it's particles move faster and gets mixed with air and smell is coming from far away

8_murik_8 [283]3 years ago
8 0
This is because the diffusion, this phenomenon occurs usually when the temperature increases, it create heat-induced motion of the higher molecules.
You might be interested in
The aromatic hydrocarbon cymene (C10H14) is found in nearly 100 spices and fragrances, including coriander, anise, and thyme. Th
andrezito [222]

Answer : The thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole

Explanation :

First we have to calculate the heat released by the combustion.

q=c\times \Delta T

where,

q = heat released = ?

c = specific heat of calorimeter = 3.640kJ/^oC

\Delta T = change in temperature = 19.35^oC

Now put all the given values in the above formula, we get:

q=3.640kJ/^oC\times (19.35)^oC

q=70.43kJ

Thus, the heat released by the combustion = 70.43 kJ

Now we have to calculate the molar enthalpy combustion.

\Delta H=-\frac{q}{n}

where,

\Delta H = molar enthalpy combustion = ?

q = heat released = 70.43 kJ

n = number of moles cymene = \frac{\text{Mass of cymene}}{\text{Molar mass of cymene}}=\frac{1.608g}{164.22g/mol}=0.009792mole

\Delta H=-\frac{70.43kJ}{0.009792mole}=-7193kJ/mole

Therefore, the thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole

4 0
3 years ago
1. Calculate the molarity when 403 g MgSO4 is dissolved to make a 5.25 L solution. Round to two significant digits.
lisov135 [29]

Answer:

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

4. Molarity of NaCl= 2.3 M

5. Mass of C₁₂H₂₂O₁₁ = 4.1 g

6. Mass of C₁₀H₈ (naphthalene) = 77 g

7. mass of ethanol = 11 g

8. Mass of DDT = 0.011 mg

Explanation:

Ans 1.

Data given

mass of MgSO₄ = 403

Volume of solution = 5.25 L

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of MgSO₄ = 24 + 32 + 4(16)

Molar mass of MgSO₄ = 120 g/mol

Put values in equation 1

             no. of moles = 403 g / 120 g/mol

             no. of moles = 3.36 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 3.36 mol / 5.25 L

           M = 0.64

So, round the figure to two significant digits.

Molarity of MgSO₄ = 0.64 M

_____________________

Ans 2.

Data given

mass of AgNO₃ = 1.24 g

Volume of solution = 125 mL

convert mL to L

1000 mL = 1 L

125 mL = 125/1000 = 0.125

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of AgNO₃ = 108 + 14 + 3(16)

Molar mass of AgNO₃ = 170 g/mol

Put values in equation 1

             no. of moles = 1.24 g / 170 g/mol

             no. of moles = 0.0073 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 0.0073 mol / 0.125 L

           M = 0.06

So, round the figure to two significant digits.

Molarity of AgNO₃ = 0.06 M

______________________

Ans 3.

Data Given:

volume of acetic acid = 500 mL

% solution of acetic acid (M)= 50%

volume of acetic acid needed = ?

Solution:

formula used

    percent of solution = volume of solute/ volume of solution x 100

Put Values in above formula

                        50 % = volume of solute / 500 mL x 100

Rearrange the above equation

                   volume of solute =   50 x 500 mL /100

                   volume of solute =   250 mL

So, volume of acetic acid = 250 mL

______________________

Ans 4

Data Given:

Molarity of NaCl (M1) = 6 M

Volume of NaCl (V1) = 750 mL

convert mL to L

1000 ml = 1 L

750 ml = 750/1000 = 0.75 L

Volume of dilute solution (V2)= 2 L

Molarity of dilute solution (M2) = ?

Solution:

Dilution Formula will be used

                M1V1 = M2V2 . . . . . . (1)

Put values in equation 1

               6  M x 0.75 L = M2 x 2 L

Rearrange the above equation

               M2 = 6 M x 0.75 L / 2 L

               M2 = 2.25 M

So, round the figure to two significant digits.

Molarity of NaCl= 2.3 M

_________________________

Ans 5.

Data Given:

Molarity of C₁₂H₂₂O₁₁ = 0.16 M

Mass of C₁₂H₂₂O₁₁ (m) = ?

Volume of dilute solution = 75 mL

convert mL to L

1000 ml = 1 L

75 ml = 75/1000 = 0.075 L

Solution:

As we know

            Molarity = no.of moles/ liter of solution . . . . . . .(1)

we also know that

           no.of moles = mass in grams / molar mass . . . . . (2)

Combine both equation 1 and 2

            Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

Put values in equation in equation 3

              0.16 M = (mass in grams / 342 g/mol) / 0.075 L

Rearrange the above equation

         mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol

         mass in grams = 4.104 g

So, round the figure to two significant digits.

Mass of C₁₂H₂₂O₁₁ = 4.1 g

____________________

The remaing portion is in attachment

8 0
4 years ago
In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to
Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

Explanation:

The measured pH of the solution = 5.153

C_6H_5OH\rightarrow C_6H_5O^-+H^+

Initially      c

At eq'm   c-x       x  x

The expression of dissociation constant is given as:

K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}

Concentration of phenoxide ions and hydrogen ions are equal to x.

pH=-\log[x]

5.153=-\log[x]

x=7.03\times 10^{-6} M

K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}

K_a=9.34\times 10^{-11}

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

4 0
4 years ago
Balance <br> Si4H10+O2&gt;SiO2+H2O
Illusion [34]

Explanation: Si4H10+O2>SiO2+H2O

2Si4H10 + 13O2 = 8SiO2 + 10H2O

5 0
3 years ago
Which is a chemical property of hydrogen?
larisa86 [58]

Answer:

<em>Option</em><em> b</em><em>)</em><em> </em><em>It </em><em>is </em><em>a </em><em>combustible</em><em> </em><em>gas </em><em>.</em>

<em><u>maybe</u></em><em><u> this</u></em><em><u> </u></em><em><u>might</u></em><em><u> help</u></em><em><u> u</u></em>

3 0
3 years ago
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