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4 years ago

Verdadero o falso: Dos sustancias en estado gaseoso(O2, N2), presentan igual

Chemistry

1 answer:

DochEvi [55]4 years ago

7 0

Srry I can’t help u because I can’t understand ur English I don’t speak French Srry I’m learning French in school

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What role does the vulture playing this food web

musickatia [10]

Answer:

scavenger- eats dead carcasses

Explanation:

4 0

3 years ago

How long would it take Jesse, with an acceleration of -2.50 m/s², to bring his bicycle, with an initial velocity of 13.5 m/s, to

lbvjy [14]

Answer:

i am so sorry. i do not have a answer but i am trying to find questioms i can answer

6 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

Name the homologous series for this compound and show its structure<br>ch3ch2cc(ch2)2ch3

kotykmax [81]

Answer:

Alkyne

Explanation:

The homologous series for this compound is the Alkyne because of the presence of the triple bond present between the third and fourth carbon atoms as seen in the picture attached.

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3 years ago

Why is oxygen diflouride a gas at room temperature?

atroni [7]

Answer:

Oxygen is a simple molecular structure, where individual oxygen atoms are bonded to each other by strong covalent bonds. Hence, a low amount of energy is required to overcome these weak forces and oxygen has a low boiling point. Therefore, at room temperature, oxygen is a gas. Oxygen difluoride is a colorless gas, condensable to a pale yellow liquid, with a slightly irritating odor. It is the most stable of the compounds of fluorine and oxygen, which include O,F,, O,F, and 0,F2 but nevertheless it is a strong oxidizing and fluorinating agent. Oxygen Difluoride is a colorless gas or a yellowish-brown liquid with a foul odor. Just to finally link Joseph's answer to the question, oxygen difluoride will thus change from liquid to solid state when chilled from -220°c to -230°c. The boiling point of oxygen is -182.96 degrees Celsius (under 1 standard atmosphere). This means at temperatures below that point, oxygen is a solid or a liquid, and at temperatures above that point, oxygen is a gas. So at -183 degrees Celsius, oxygen is a liquid.

Explanation:

3 0

3 years ago