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3 years ago

How much energy is needed to completely boil a 5.05g sample of water?

1 answer:

8 0

Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

<h3>Why does it take this much energy to boil the water?</h3>

We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.

Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

To learn more about the behavior of water visit:

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The rate of decay of a chemical involved in a reaction that is second order (bimolecular) in one reactant A is given by: = k [AJ

anyanavicka [17]

Explanation:

$2A\rightarrow products$

According to mass action,

$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2$

Where, k is the rate constant

So,

$\dfrac{d[A]}{dt}=-k[A]^2$

Integrating and applying limits,

$\int_{[A_t]}^{[A_0]}\frac{d[A]}{[A]^2}=-\int_{0}^{t}kdt$

we get:

$\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Half life is the time when the concentration reduced to half.

So, $[A_t]=\frac{1}{2}\times [A_0]$

Applying in the equation as:

$t_{1/2}=\dfrac{1}{k[A_o]}$

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4 years ago

Biochemistry is the study of the proceses that take place in...

weeeeeb [17]

Answer:

Living organisms/chemical processes that support life

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3 years ago

what is the total number of grams of NaOH needed to react completely with 196 grams of H2 SO4? regents

8090 [49]

Answer:

160 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $H_2SO_4$ :-

Mass of $H_2SO_4$ = 196 g

Molar mass of $H_2SO_4$ = 98 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{196\ g}{98\ g/mol}$

$Moles\ of\ Sulfuric\ acid= 2\ mol$

According to the given reaction:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

1 mole of sulfuric acid reacts with 2 moles of NaOH

So,

2 moles of sulfuric acid reacts with 2*2 moles of NaOH

Moles of NaOH must react = 4 moles

Molar mass of NaOH = 40 g/mol

<u>Mass = Moles*molar mass = $4\times 40\ g$ = 160 g</u>

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4 years ago

What independent variables can be tested in this lab setup? (List at least 5)

tigry1 [53]

Answer:

I know 2

dependent

independent

Explanation:

sorry that's about it from me

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3 years ago

For the following equilibrium system, which of the following changes will form more CaCO3? CO2(g) + Ca(OH)2(s) ⇌ CaCO3(s) + H2O(

Talja [164]

Answer: d. Remove one-half of the initial CaCO3.

Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:

A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;

A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.

A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.

So, the correct answer is D. Remove one-half of the initial CaCO3.

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4 years ago