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3 years ago

How long do you leave hydrogen peroxide in your ear

Chemistry

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

2 mins

Explanation:

because it will let it soak into your ear and get everything out

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An emulsifying agent is typically characterized by having

pochemuha

The third substance or agent which produce the film between the interface of two immiscible liquids and thus stabilize the system are known as an emulsifying agent.

Since the solubility of the liquids depends on the polarity of the mixing liquids the thumb rule of solubility is like dissolves like that means polar liquid dissolves in polar liquid only and vice versa. For two immiscible liquids, the emulsifying agent is used which does not chemically change the polarity of liquids but acts as bridge between immiscible liquids, the polar end of the emulsifier attach to the polar liquid and the non-polar end of the emulsifier attach to the non-polar end and thus help in dissolving.

Therefore, the one end of the emulsifier is polar and the other end is non-polar

5 0

4 years ago

¿Que hace diferente ala química de la alquimia?

viva [34]

Answer:

La distinción entre la química teóricamente teórica y la química avanzada es la química teórica teórica que se basa en un espiritualista, potente que percibe visualmente la autenticidad, mientras que la química espera que la autenticidad sea fundamentalmente mundana. Eso engendra una tremenda distinción, y la química nunca se hubiera alejado excepcionalmente de la remota posibilidad de que se hubiera quedado con el trascendentalismo antediluviano.

Explanation:

8 0

3 years ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

4 years ago

How can you overcome drawback of a chemaicl equation ? any three

Shtirlitz [24]

Answer:

1.rate of reaction should be indicated by symbols 2.we should indicate physical states by sybloms 3.reversible reaction should be indicated by using double arrows

4 0

3 years ago

1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

We have the following data:

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:

$P_0*V_0 = P*V$