3 years ago

Sometimes the oxidation state of an analyte must be adjusted before it can be titrated.

Chemistry

1 answer:

alexdok [17]3 years ago

5 0

Answer:

Explanation:

True

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A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re

sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature 10.0°C + 273 = 283 K

T₁ = initial temperature , 40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

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The fuel used to power the booster rockets on

Effectus [21]

Answer:

746 moles of H2O are been produced from 373 moles of Al.

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For every 3 moles of aluminum, you get 6 moles of H2O (double). Therefore, every 373 moles of Al, you will get double as well, that is 746 m.

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Which step will decrease the pressure of a gas inside a closed cubical container? increasing the number of moles of gas decreasi

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Answer: Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.

Explanation:

According to Gay-Lussac's Law : 'The pressure of the gas increases with increase in temperature of the gas when volume of the gas is kept constant'.

$(Pressure)\propto (Temperature)$

At constant volume, pressure of the gas will decrease on decreasing the temperature or vice versa.

Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.

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aleksley [76]

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