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statuscvo [17]
3 years ago
14

The process in which organisms grow and replace worn-out cells is called: A. Cell regeneration B. Cell division C. Mitosis D. Bo

th b and c
Chemistry
1 answer:
djyliett [7]3 years ago
7 0
The mitosis is the cell division so it both B and C
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Ammonium carbonate decomposes upon heating according to
BaLLatris [955]

The total volume of the gas produced, given the data from the question is 11.84 L

<h3>How to determine the mole of (NH₄)₂CO₃</h3>
  • Mass of (NH₄)₂CO₃ = 11.83 g
  • Molar mass of (NH₄)₂CO₃ = 96 g/mol
  • Mole of (NH₄)₂CO₃ =?

Mole = mass / molar mass

Mole of (NH₄)₂CO₃ = 11.83 / 96

Mole of (NH₄)₂CO₃ = 0.123 mole

<h3>How to determine the volume of NH₃</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 2 moles of NH₃

Therefore,

0.123 mole of (NH₄)₂CO₃ will decompose to produce = 0.123 × 2 = 0.246 mole of NH₃

Thus, we can determine the volume of NH₃ produced by using the ideal gas equation as follow:

  • Temperature (T) = 22 °C = 22 + 273 = 295 K
  • Pressure (P) = 1.02 bar
  • Gas constant (R) = 0.08314 bar.L/Kmol
  • Number of mole (n) = 0.246 moles
  • Volume of NH₃ (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.246 × 0.08314 × 295) / 1.02

Volume of NH₃ = 5.92 L

<h3>How to determine the volume of CO₂</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 1 moles of CO₂

Therefore,

0.123 mole of (NH₄)₂CO₃ will also decompose to produce = 0.123 mole of CO₂

Thus, we can determine the volume of CO₂ produced by using the ideal gas equation as follow:

  • Temperature (T) = 22 °C = 22 + 273 = 295 K
  • Pressure (P) = 1.02 bar
  • Gas constant (R) = 0.08314 bar.L/Kmol
  • Number of mole (n) = 0.123 moles
  • Volume of CO₂ (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.123 × 0.08314 × 295) / 1.02

Volume of CO₂ = 2.96 L

<h3>How to determine the volume of H₂O</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 1 moles of H₂O

Therefore,

0.123 mole of (NH₄)₂CO₃ will also decompose to produce = 0.123 mole of H₂O

Thus, we can determine the volume of H₂O produced by using the ideal gas equation as follow:

  • Temperature (T) = 22 °C = 22 + 273 = 295 K
  • Pressure (P) = 1.02 bar
  • Gas constant (R) = 0.08314 bar.L/Kmol
  • Number of mole (n) = 0.123 moles
  • Volume of H₂O (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.123 × 0.08314 × 295) / 1.02

Volume of H₂O = 2.96 L

<h3>How to determine the total volume of gas produced</h3>
  • Volume of NH₃ = 5.92 L
  • Volume of CO₂ = 2.96 L
  • Volume of H₂O = 2.96 L
  • Total volume of gas =?

Total volume = 5.92 + 2.96 + 2.96

Total volume of gas = 11.84 L

Learn more about ideal gas equation:

brainly.com/question/4147359

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

7 0
2 years ago
Which sentence from the article BEST describes HOW fireworks produce colors in the sky?
aleksley [76]
The answer is Abecaue it is the most specific on how
7 0
3 years ago
How many moles are in 175.9 grams of Pb3(PO4)2?
snow_lady [41]

Answer:

0.21674767727138733

Explanation:

Again I don't want to give an explanation right now because I can't exactly do that more than once. it hurts my child brain. And the fact that I just woke up.

8 0
3 years ago
At what pH are the moles of acid and base equal?
mihalych1998 [28]

Answer:

7

Explanation:

(Note: for a strong acid and strong base titration the equivalence point is at a pH=7. This is because at this point you have equal moles of added base as acid in the original solution. Therefore at the equivalence point the solution has formed a neutral salt and the pH is 7).

5 0
3 years ago
Read 2 more answers
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

5 0
3 years ago
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