Ask question

Ask question

All categories

Montano1993 [528]

3 years ago

11

El agua oxigenada se descompone rápidamente en agua y oxígeno cuando se le agrega una peueña cantidad de dióxido de manganeso, q

ue no se altera durante la reacción y se recupera totalmente al finalizar la misma ¿Cómo se comporta el dióxido de manganeso en esta reacción?

1 answer:

katovenus [111]3 years ago

6 0

Answer:

IDK

Explanation:

I dont speak spanish

You might be interested in

You mix a blue and a clear liquid together and they turn bright green. Has a chemical reaction occurred?

Margarita [4]

Yes because color change is a sign of a chemical reaction.

7 0

3 years ago

Read 2 more answers

Please help me need it fast

Schach [20]

Answer:

Do you need 3 ways or just one?

1. Temperature.

2. Pressure.

3. Polarity.

Explanation:

Eh hope these help, Idr understand the question but those are 3 ways to increase the solubility of a solid in water.

5 0

2 years ago

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper

Olin [163]

Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

= (M)(Volume of the solution (L))

= (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

= Qrxn/(no. of moles of AgNO₃)

= (1170.4 J)/(0.03 mol)

= 39013.33 J/mol

= 39.013 kJ/mol.

7 0

3 years ago

(GIVING BRAINLYIST) How does temperature and precipitation change as you increase in elevation?

S_A_V [24]

Answer:

Temperature gets cooler and precipitation increases.

Explanation:

As average temperatures at the Earth's surface rise, more evaporation occurs, which, in turn, increases overall precipitation. … In addition, higher temperatures lead to more evaporation, so increased precipitation will not necessarily increase the amount of water available for drinking, irrigation, and industry

Higher elevations cause temperatures to drop because the higher up in the atmosphere you go, the colder it gets due to air pressure, in turn causing precipitation to freeze, creating snow. Lower elevations are a lot warmer because the air pressure is not as pressurized causing the temperature to be warmer

7 0

2 years ago

Read 2 more answers

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago