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3 years ago

Qual é a etapa onde ocorre a adição de cloro para eliminação de microorganismos patogênicos? AULA DE CIÊNCIAS 6° ano

Chemistry

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

A desinfecção

Explanation:

A desinfecção é o processo no qual o cloro é adicionado para matar microorganismos patogênicos prejudiciais, como bactérias, vírus, etc. Este é um processo químico e biológico. alguns desinfetantes comuns são cloro, dióxido de cloro e ozônio.

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If 2 objects have the same mass of 100 grams and object A has a volume of 50cm 3 and object B has a volume of 200 cm 3, which of

kenny6666 [7]

Answer:

The object A will be having the greater density compared to object B.

Explanation:

It is known that density of any object is defined as the mass of any object occupying a given volume. So the ratio of mass and volume will help to determine the density of any object. $Density=Mass/Volume$

From the above equation, it can be seen that the density of any object is directly proportional to the mass of the object and inversely proportional to the volume occupied by the object.

So in the present context, the mass of objects A and B are same and it is 100 g. Thus, the density of object A and object B will be influenced by their volume. As it is given that the volume of object A is 50 cm3 and object B is 100 cm3, then depending upon the relationship of volume and density, the density of both the objects can be determined. As the object with higher volume will be having lesser density as volume is inversely proportional to density. Thus, in the given case the volume of object B is greater than object A and so the object A will be having greater density compared to object B.

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3 years ago

The table below shows two types of electromagnetic waves and three random applications of electromagnetic waves.

Reptile [31]

Correct Answer is 1 i.e. Gamma rays—2 and radio waves—3

Reason:
1) In a hypernova, star as similar to nuclear fusion converts lighter elements into heavy elements. If fusion is not capable of generating enough pressure to counteract gravity, star immediately collapses to form a black hole. During this process, energy will be released, along the axis of rotation to form gamma-ray burst. Such gamma-ray burst was first detected using Fermi Gamma-ray Space Telescope. Thus, gamma-ray is capable of providing information of gravity fields.

2) Radiowaves are capable of inducing transitions that requires less energies. These transition includes nuclear excitation and electron excitation (in rotational energy level). Depending upon the value to Jmax, it is possible to determine the temperature and heat released by astronomical objects

4 0

3 years ago

Read 2 more answers

i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i

9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

$C_8H_{18}+O_2\to CO_2+H_2O$

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

$C_8H_{18}+O_2\to8CO_2+H_2O$

Now, we balance the hydrogen:

$C_8H_{18}+O_2\to8CO_2+9H_2O$

And in the end, the oxygen:

$C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O$

We can multiply all coefficients by 2 to get integer ones:

$2C_8H_{18}+25O_2\to16CO_2+18H_2O$

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

$\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}$

For the products, we have:

$\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}$

Now, we substract the rectants from the produtcs:

$\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}$

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

$\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ$

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

$\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}$

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

$Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ$

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, m.

First, we have to heat the ice to 0 °C, so:

$\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}$

Then, we need to melt all this mass, so we use the latent heat now:

$Q_2=n\cdot6.03kJ/mol$

Converting mass to number of moles of water we have:

$\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}$

So:

$Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g$

Adding them, we have a total heat of:

$\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}$

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):