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Maksim231197 [3]
3 years ago
12

Three bags containing different number of wooden cubes are dropped from the same height. Which of the following statements is tr

ue?
A) Bag A has more kinetic energy than Bag C as it falls
B) Bag C has the most kinetic energy before it was release
C) Bag B has less kinetic energy than Bag A as it falls.
D) Bag A has the least potential energy when it is dropped
Chemistry
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

I don't have the number of cubes in each bag, but whichever bag had the most cubes would have the most kinetic energy as it falls

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Assuming the partial pressure of oxygen in air (0.20 atm) and nitrogen in air (0.80 atm). Calculate the mole fractions of oxygen
Nuetrik [128]

Answer:

oxygen = 4.7 * 10^-6

Nitrogen = = 9.7 * 10^6

Explanation:

partial pressure of oxygen = 0.20 atm

partial pressure of Nitrogen = 0.80 atm

<u>calculate the mole fractions of oxygen and Nitrogen in water </u>

Temp = 298k

applying henry's law

molar conc of oxygen in water ( Coxygen )

= Kp = 1.3 * 10^-3 Mol/L.atm * 0.20 atm = 2.6 * 10^-4 Mol

molar conc of Nitrogen in water ( Cnitrogen )

= Kp = 6.8 * 10^-4 Mol/L.atm * 0.80 atm = 5.4 * 10^-4

next Given that the number of moles in 1 liter of water = 55.5 mol  

therefore the mole fraction of oxygen

= 2.6 * 10^-4 / 55.5

= 4.7 * 10^-6

mole fraction of Nitrogen

= 5.4 * 10^-4  / 55.5

= 9.7 * 10^6

6 0
2 years ago
A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the
Stells [14]

Answer:

e. T₂= 4T₁

Explanation:

Initially, we have a number of moles (n₁) a gas sample at a certain pressure (P), temperature (T₁) and volume (V). We can relate these variables through the ideal gas equation.

P . V = n₁ . R . T₁

where,

R is the ideal gas constant

We can rearrange this equation like:

T_{1}=\frac{P.V}{n_{1}.R}

If only one fourth of the initial molecules remain n₂ = 1/4 n₁. The new temperature (T₂) assuming pressure and temperature remain constant is:

P.V=n_{2}.R.T_{2}=\frac{1}{4} n_{1}.R.T_{2}\\\frac{P.V}{n_{1}.R} =\frac{1}{4} T_{2}\\T_{1}=\frac{1}{4} T_{2}\\T_{2}=4.T_{1}

6 0
3 years ago
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Vinil7 [7]

Answer:

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Explanation:

4 0
3 years ago
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Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t
Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of HC_2H_3O_2 \  (M_1) = 0.105 M

Volume of  HC_2H_3O_2 \  (V_1) = 20.0 mL

Concentration of NaOH (M_2) = 0.125 M

The  chemical reaction can be expressed as:

HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}

Using the ICE Table to determine the equilibrium concentrations.

          HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}

I            0.105                                     0                  0

C              -x                                         +x                +x

E            0.105 - x                                  x                  x

K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}

K_a = \dfrac{(x)(x)}{(0.105-x)}

Recall that the ka for HC_2H_3O_2= 1.8 \times 10^{-5}

Then;

1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}

1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}

By solving the above mathematical expression;

x = 0.00137 M

H_3O^+ = x = 0.00137  \ M \\ \\  pH = - log [H_3O^+]  \\ \\  pH = - log ( 0.00137 )

pH = 2.86

Hence, the initial pH = 2.86

b)  To determine the volume of the added base needed to reach the equivalence point by using the formula:

M_1 V_1 = M_2 V_2

V_2= \dfrac{M_1V_1}{M_2}

V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}

V_2 = 16.8 mL

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of HC_2H_3O_2 = molarity × volume

= 0.105  \ M \times 20.0 \times 10^{-3} \ L

= 2.1 \times 10^{-3}

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = 0.625 \times 10^{-3}

After reacting with 5.0 mL NaOH, the number of moles is as follows:

                    HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}

Initial moles   2.1*10^{-3}       0.625 * 10^{-3}           0                      0

F(moles) (2.1*10^{-3} - 0.625 \times 10^{-3})    0      0.625 \times 10^{-3}         0.625 \times 10^{-3}

The pH of the solution is then calculated as follows:

pH = pKa + log \dfrac{[base]} {[acid]}

Recall that:

pKa for HC_2H_3O_2=4.74

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the (C_2H_3O^-_2) with H_2O

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of (C_2H_3O^-_2) = moles/volume

= \dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}

= 0.0571 M

Now, using the ICE table to determine the concentration of H_3O^+;

             C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}

I              0.0571                                0                      0

C              -x                                       +x                     +x

E             0.0571 - x                             x                       x

Recall that the Ka for HC_2H_3O_2 = 1.8 \times 10^{-5}

K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} }  \\ \\ K_b = 5.6 \times 10^{-10}

k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}

5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}

x = [OH^-] = 5.6 \times 10^{-6} \ M

[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }

[H_3O^+] =1.77 \times 10^{-9}

pH =-log  [H_3O^+]   \\ \\  pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

             HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}

Before                             0.0021              0.002725         0

After                                   0                     0.000625        0.0021

[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 )  \ L}

[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}

[OH^-] =  0.0149 \ M

From above; we can determine the concentration of H_3O^+ by using the following method:

[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}

[H_3O^+] = 6.7 \times 10^{-13}

pH = - log [H_3O^+]

pH = -log (6.7 \times 10^{-13} )

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

3 0
3 years ago
How many cobalt atoms are in 2 moles of cobalt?
Hunter-Best [27]

That means that one mole of cobalt weighs 58.9332 grams (58.9332 g/mol). Based on that information, to convert 2 moles of cobalt to grams, we multiply 2 moles of cobalt by 58.9332.

4 0
2 years ago
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