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2 years ago

Why is it important to note the temperature when determining the density of a substance?

1 answer:

7 0

It is important to take note of th temperature in determining the density of a substance because this will set as a basis and will likely be a variable in the experiment because this will also contribute on the effects of the experiment and a basis of how the experiment has turned to be that way.

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Which enzyme is responsible for facilitating the hydrogen bonding

jeka57 [31]

Answer:

its the polymerase

5 0

3 years ago

2. Calculate the atomic mass of an element that has two isotopes, each with 50.00% abundance. One isotope has a mass of 63.00 am

melamori03 [73]

Answer:

The atomic mass of element is 65.5 amu.

Explanation:

Given data:

Abundance of X-63 = 50.000%

Atomic mass of X-63 = 63.00 amu

Atomic mass of X-68 = 68.00 amu

Atomic mass of element = ?

Solution:

Abundance of X-68 = 100-50 = 50%

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (50×63)+(50×68) /100

Average atomic mass = 3150 + 3400 / 100

Average atomic mass = 6550 / 100

Average atomic mass = 65.5 amu.

The atomic mass of element is 65.5 amu.

7 0

2 years ago

Which of these statements best describes how an increase in the amount of Sun’s heat reflected by the Earth’s surface and atmosp

Korolek [52]

If the Suns heat is being reflected, then it is making the Earth cooler.

7 0

3 years ago

The oxidation of Cu₂O(s) to CuO(s) is an exothermic process:

Alex73 [517]

Answer:

w = 7376.6 J

Explanation:

To calculate the work done in the system, we need first to calculate the number of moles of all compounds.

The following expression, that comes from the 1° law of thermodynamic, will help you to get the work:

w = ΔnRT

Where:

Δn: difference in the number of moles between products and reactants

R: constant universal of gases (8.314 J / mol K)

T: Temperature in Kelvin.

As we are doing this reaction in STP, then the Pressure is 1 atm, and temperature is 0 °C or 273 K.

Now, we already have the moles of Cu₂O, let's see, according to the balanced reaction, how many moles we should have of O₂ and CuO.

If 1 mole of O₂ reacts with 2 moles of Cu₂O then 6.5 moles of Cu₂O will be:

moles O₂ = 6.5 moles Cu₂O * (1 mole O₂/2 moles Cu₂O) = 3.25 moles O₂

If 2 moles of Cu₂O produces 4 moles of CuO, then 6.5 moles will be:

moles CuO = 6.5 moles Cu₂O*(4 moles CuO/2 moles Cu₂O) = 13 moles CuO

Now that we have the moles, let's calculate the value of Δn.

Δn = moles product - moles reactants

Δn = 13 - (6.5 + 3.25) = 3.25 moles

Now, we can calculate the work done in the system:

w = 3.25 * (8.314 * 273)

Hope this helps

8 0

2 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

3 years ago