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2 years ago

3c reaction time =138.5 sec reation rate=

Chemistry

1 answer:

blsea [12.9K]2 years ago

7 0

Answer:

Compute the reaction rate to the nearest whole number using the formula below,

mass of tablet/volume of water

Reaction Rate = mas

reaction time

3°C Reaction time = 138.5 sec

Reaction rate = mg/l/sec

24°C Reaction time = 34,2 sec

Reaction rate = mg/L/sec

40°C Reaction time = 26.3 sec

Reaction rate = mg/L/sec

65°C Reaction time = 14.2 sec

Reaction rate = mg/L/sec

Explanation:

I hope my answer help

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Consider dissolving sugar in water and magnesium sulfate in water. is any new substance being produced? what happens after sodiu

Olenka [21]

Answer:

Here's what I get

Explanation:

1. Sugar

(a) Dissolving in water

The white solid dissolves in water to give a colourless solution. There is no evidence that a new substance is being produced.

(b) Addition of sodium hydroxide

Adding the colourless solution of sodium hydroxide to the colourless sugar solution gives a colourless solution. There is no evidence that a new substance is being produced.

2. Magnesium sulfate

(a) Dissolving in water

The colourless crystals dissolve in water to give a colourless solution. There is no evidence that a new substance is being produced.

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Adding the colourless solution of sodium hydroxide to the colourless solution of magnesium sulfate gives a white precipitate (see image). This is evidence that a new substance is being produced.

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3 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Answer:

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3 years ago

Consider this reaction: HCO3− + H2S → H2CO3 + HS− Which is the Bronsted-Lowry base? H2S HCO3- HS– H2CO3

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Answer:

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