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2 years ago

Why do you think steeples and roofs might appear green, but a copper square does not?

1 answer:

8 0

Because the copper roofs, have been outside for a while and oxidization has occurred.
When metals become oxidized they rust, but when copper becomes oxidized, it turns greenish.
So copper roofs oxidize and turn green from being outside and a copper square doesn’t if it’s inside.

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How does the process of cellular respiration create energy?

Pavlova-9 [17]

Answer:

well, through the process of cellular respiration, the energy in food is converted into energy that can be used by the body's cells. During cellular respiration, glucose and oxygen are converted into carbon dioxide and water, and the energy is transferred to ATP

7 0

2 years ago

Read 2 more answers

Enter your answer in the provided box. Calcium hydroxide may be used to neutralize (completely react with) aqueous hydrochloric

NNADVOKAT [17]

Answer:

49.95 g of HCl

Explanation:

Let's formulate the chemical equation involved in the process:

Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O

This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2

1 mole Ca(OH)2 ---- 2 moles HCl

0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl

Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:

1 mole of HCl ---- 36.46094 g

1.37 moles ------ x = 49.95 g of HCl

3 0

3 years ago

Identify the compound with the smallest dipole moment in the gas phase. identify the compound with the smallest dipole moment in

stepladder [879]

Without solving for the dipole moment, we can easily determine which among the common gases has the smallest dipole moment just by determining the differences in their electronegativity. The greater the difference in the electronegativity, the higher is the value of the dipole moment.

From the given above, there are obvious differences between the electronegativity between the atoms composing LiF, ClF, and HF. For Cl2, since this is the same molecule then, the difference in the electronegativity is zero.

Answer: Cl2.

3 0

3 years ago

31. calculate the amount of energy released as heat by the freezing of 13.3 g of a liquid substance, given that the substance ha

spin [16.1K]

From the calculations, the heat of fusion of the substance is 0.73 kJ

<h3>What is is the heat of freezing?</h3>

The heat of freezing is the energy released when the substance is converted from liquid to solid.

Now we know that the molar mass of the substance is 82.9 g/mol hence the number of moles of the substance is; 13.3 g /82.9 g/mol = 0.16 moles

Now the heat of fusion shall be;

H = 4.60 kj/mol * 0.16 moles

H = 0.73 kJ

Learn more about freezing:brainly.com/question/3121416

#SPJ4

5 0

1 year 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