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2 years ago

What is a water cycle

Chemistry

2 answers:

Rina8888 [55]2 years ago

5 0

Answer:

it shows the continuous movement of water within the earth and atmosphere

marissa [1.9K]2 years ago

4 0

Answer:

the cycle of water and how it evaporates with energy from the sun and the droplets are stored in clouds which is condensation and when it rains or snows those are forms of precipitation

Explanation:

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Fill in the coefficients that will balance the following reaction: a0Ca + a1CO2 + a2O2 → a3CaCO3

VashaNatasha [74]

The chemical equation without coefficients is:

Ca + CO2 + O2 --------> Ca CO3

You can balance that equation by trial an error.

This is the chemical equation balanced:

2Ca + 2CO2 + O2 --------> 2Ca CO3

Count the atoms on each side to check the balance

Atom Left side    right side

Ca    2    2

C    2    2

O 2*2 + 2 = 6    2*3 = 6

Then those are the coefficients:

a0 = 2

a1 = 2

a2 = 1

a3 = 2

6 0

4 years ago

If 3.53 g of CuNO, is dissolved in water to make a 0.330 M solution, what is the volume of the solution in milliliters?

solmaris [256]

Answer:

84.8 mL

Explanation:

From the question given above, the following data were obtained:

Mass of CuNO₃ = 3.53 g

Molarity of CuNO₃ = 0.330 M

Volume of solution =?

Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:

Mass of CuNO₃ = 3.53 g

Molar mass of CuNO₃ = 63.5 + 14 + (16×3)

= 63.5 + 14 + 48

= 125.5 g/mol

Mole of CuNO₃ =?

Mole = mass / Molar mass

Mole of CuNO₃ = 3.53 / 125.5

Mole of CuNO₃ = 0.028 moles

Next, we shall determine the volume of the solution. This can be obtained as follow:

Molarity of CuNO₃ = 0.330 M

Mole of CuNO₃ = 0.028 moles

Volume of solution =?

Molarity = mole /Volume

0.330 = 0.028 / Volume

Cross multiply

0.330 × Volume = 0.028

Divide both side by 0.330

Volume = 0.028 / 0.330

Volume = 0.0848 L

Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.0848 L = 0.0848 L × 1000 mL / 1 L

0.0848 L = 84.8 mL

Therefore, the volume of the solution is 84.8 mL.

8 0

3 years ago

Are there molecules of sodium chloride within a sodium chloride crystal?

alukav5142 [94]

No, within a crystal like structure or ionic lattice of sodium chloride, there are ions of Na and Cl. Na+ and Cl- respectively that are attracted to each other due to their opposite charges. Many of these ions form a crystal structure.

6 0

4 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

4 years ago

How many atoms are there in 4 moles of table sugar ?

Likurg_2 [28]

Answer: 2.408×10^24 atoms

Explanation:

Applying the formula

No. moles= (no. Of atoms)/Avogadro's number

4 = n/6.02×10^23

no. of atoms= 2.408×10^24 atoms

7 0

3 years ago

What is a water cycle​

What is a water cycle