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Luden [163]
3 years ago
8

We must __________ forces to calculate net force if the forces are going in OPPOSITE directions.

Chemistry
1 answer:
Virty [35]3 years ago
5 0

Answer:

subtract

Explanation:

To find the net force of two bodies going in opposite directions, we must subtract the forces this is because they are not additive forces.

  • The net force is the resultant force acting on a body.
  • This resultant force is the single force that will produce the same effects as all given forces.
  • When forces are directed oppositely on a body, they are subtracted from each other to find the net force.
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According to this graph, what is the difference in average rainfall from October to November? (Choose the best answer)
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Answer:its 1

Explanation:

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3 years ago
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What is the electron configuration for iodine
iogann1982 [59]

Hello!

To find electron configuration for Idoine we need to understand the following steps:

  1. Finding the Atom's Atomic Number (tells us the specific number of electrons)
  2. Determining the Charge of the Atom
  3. Understanding the orbitals (Set S [Contains 2], P [Contains 3, Holds 6], D [Contains 5, Holds 10], F [Contains 7, Holds 14], and there are some theoritical ones.) [Overall the sets go SPDFGHIK
  4. Understanding notations in configuartion. The notations display the number of electrons in the atom and set.

In this case, for Iodine. If we follow these rules we can see that the electron configuration is [Kr] 4d^10 5s^2 5p^5. We use Krytpon in front because that is the last full and stable noble gas before this particular element. Atoms are just trying to be stable so the goal is to achieve that full shell.

7 0
3 years ago
Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at
ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and O_2.

\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

4Al+3O_2\rightarrow 2Al_2O_3

From the balanced reaction we conclude that

As, 3 mole of O_2 react with 4 mole of Al

So, 0.188 moles of O_2 react with \frac{4}{3}\times 0.188=0.251 moles of Al

From this we conclude that, Al is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Al_2O_3

From the reaction, we conclude that

As, 3 mole of O_2 react to give 2 mole of Al_2O_3

So, 0.188 moles of O_2 react to give \frac{2}{3}\times 0.188=0.125 moles of Al_2O_3

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of Al_2O_3 releases amount of heat = 1676 kJ

So, 0.125 mole of Al_2O_3 releases amount of heat = 0.125\times 1676kJ=209.5kJ

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0
3 years ago
Estimate ΔG°rxn for the following reaction at 775 K. 2Hg(g) + O2(g) → 2HgO(s) ΔH°= -304.2 kJ; ΔS° = -414.2 J/K
hichkok12 [17]

Answer:

\Delta G^0_{rxn}=16.81kJ

Explanation:

Hello!

In this case, since the Gibbs free energy of any process is related with the enthalpy change, temperature and entropy change as shown below:

\Delta G^0=\Delta H^0-T\Delta S^0

For a chemical reaction it is simply modified to:

\Delta G^0_{rxn}=\Delta H^0_{rxn}-T\Delta S^0_{rxn}

Thus, since the enthalpy of reaction is given as -304.2 kJ and the entropy as -414.2 J/K (-0.4142 kJ/K), at 775 K the Gibbs free energy of reaction turns out:

\Delta G^0_{rxn}=-304kJ-775K*(-0.4142\frac{kJ}{K} )\\\\\Delta G^0_{rxn}=16.81kJ

Whose result means this is a nonspontaneous reaction.

Best regards!

3 0
3 years ago
Acetylating agents such as acetic anhydride react preferentially with primary amines, iodoacetate reacts preferentially with sul
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II think thihe answer is to be part of the acid to be honest

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