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Agata [3.3K]
3 years ago
11

When Truman walks across the tile floor in his kitchen, the sound of his shoes is loud and sharp. Which of the following best de

scribes what would happen if he walked over carpet in the same shoes?
Chemistry
1 answer:
Alexxx [7]3 years ago
6 0

Answer:

Soft and quiet

Explanation:

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consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
iogann1982 [59]

Answer:

pH=10.5

Explanation:

Hello,

In this case, the dissociation of the given weak acid is:

HA\rightleftharpoons H^++A^-

Therefore, the law of mass action for it turns out:

Ka=\frac{[H^+][A^-]}{[HA]}

That in terms of the change x due to the reaction extent is:

1x10^{-23}=\frac{x*x}{0.1-x}

Thus, by solving with the quadratic equation or solver, we obtain:

x=31.6x10^{-12}M

Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:

pH=-log(-31.6x10^{-12})\\pH=10.5

Regards.

5 0
3 years ago
A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper
finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0
3 years ago
Which reacts faster, rock salt or grains of salt? Why?​
sweet-ann [11.9K]

Answer:

salt

Explanation:

because it has a much larger volume ratio

5 0
3 years ago
There is no connection between the study of chemistry and physics. A. true B. false
9966 [12]
B. False, have a connection between chemistry and physis
4 0
3 years ago
N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
3 years ago
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