Buckminsterfullerene is an allotrope of carbon that is found in:

aluminum has a density or 2.7g/cm3. copper has a density of 8.96/cm3. which metal would you choose to build a model airplane?

erik [133]

Answer: Aluminium

Explanation: Aluminium metal has a lower density than copper. So, for the same volume of metal used to build a model airplane, the aluminium plane would be very lightweight while that of copper would be heavy. The lightweight airplane will fly easily.

7 0

3 years ago

Which sound has waves with the greatest amplitude?'

Furkat [3]

Jet Takeoff because it is the loudest according to the chart by the number of decibels.

8 0

3 years ago

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

3 years ago

When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of

taurus [48]

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

3 0

3 years ago

Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5

lubasha [3.4K]

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

$K_{a}=\frac{[H^{+}][HB] }{[reactant]}$

[HB] is the concentration of base.

$8 * 10^{-5} =\frac{x^{2} }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1$

$x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}$

Then

$pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54$

So the pH of the solution is 2.54.

4 0

3 years ago