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3 years ago

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The photograph shows the most abundant substance on Earth's su Which fact about this substance tells you that it is a compound?

A. It is a colorless liquid that is inflammable. B. It is made up of atoms that have mass and take up space c. It can be separated into two substances by a chemical pro D. It appears to be the same type of matter throughout.

2 answers:

Arlecino [84]3 years ago

5 0

Answer:

B. It is made up of atoms that have mass and take up space

rosijanka [135]3 years ago

4 0

Answer:

the answer to this question is b

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If 25.3 grams of mercury(II) oxide react to form 23.4 grams of mercury, how many grams of oxygen must simultaneously be formed?

sergij07 [2.7K]

Answer : The mass of oxygen formed must be 3.8 grams.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

$2HgO\rightarrow 2Hg+O_2$

According to the law of conservation of mass,

Total mass of reactant side = Total mass of product side

Total mass of $2HgO$ = Total mass of $2Hg+O_2$

As we are given :

The mass of $HgO$ = 25.3 grams

The mass of $Hg$ = 23.4 grams

So,

$2\times 25.3g=2\times 23.4g+\text{Mass of }O_2$

$50.6g=46.8g+\text{Mass of }O_2$

$\text{Mass of }O_2=3.8g$

Therefore, the mass of oxygen formed must be 3.8 grams.

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3 years ago

How many molecules are in 2.50 moles of carbon dioxide?

podryga [215]

Answer:

1.5055×10²⁴ molecules

Explanation:

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

Number of mole CO₂ = 2.5 moles

Number of molecules CO₂ =?

The number of molecules present in 2.5 moles CO₂ can be obtained as:

From Avogadro's hypothesis,

1 mole of CO₂ = 6.022×10²³ molecules

Therefore,

2.5 mole of CO₂ = 2.5 × 6.022×10²³

2.5 mole of CO₂ = 1.5055×10²⁴ molecules

Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂

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3 years ago

(5 simple questions)

maks197457 [2]

1)a. formation of gas
2)b.energy was transferred
3)c.substances that are used up in a reaction

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What kinds of sacrifices do governments face when considering how to reduce greenhouse gases ?

guapka [62]

Answer:

m risus ante, dapibus a ipsum dolor sit amet, add cream or a great slaughter. Fusce dui lectus, laoreet congue vel, it was said of life in hatred. Until the bananas. Lorem ipsum dolor sit amet, consectetur adipiscing elit. In fact, lacinia pulvinar dolor libero facilisis. Nutrition protein produced et. In fact, risus ante, dapibus a ipsum dolor sit amet, add cream or a great slaughter. Fusce dui lectus, laoreet congue vel, it was said of life in hatred. Until the bananas. Lorem ipsum deceit

Explanation:

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3 years ago

Read 2 more answers

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