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2 years ago

Peripherals are used to?

Chemistry

1 answer:

olga2289 [7]2 years ago

7 0

Answer:

Peripheral device, also known as peripheral, computer peripheral, input-output device, or input/output device, any of various devices (including sensors) used to enter information and instructions into a computer for storage or processing and to deliver the processed data to a human operator or, in some cases.

Explanation:

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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

2 years ago

Which compound is a polar molecule and has a linear geometry

kirill115 [55]

Answer:

Linear molecule is a molecule in which atoms are deployed in a straight line (under 180° angle). Molecules with an linear electron pair geometries have sp hybridization at the central atom. An example of linear electron pair and molecular geometry are carbon dioxide (O=C=O) and beryllium hydride BeH2.

Explanation:

6 0

3 years ago

1. What is the molarity of a solution prepared by dissolving 3.11 grams of NaOH in

Elena L [17]

The molarity of a solution prepared by dissolving 3.11 grams of NaOH in

enough water to make 300 milliliters of solution is 0. 256 mol/ L

Explanation:

<h3>What is Molarity?</h3>

Molarity (M) is the amount of a substance in a given volume of solution. It is otherwise known as the number of moles of solute in a certain amount of solution.

The unit of molarity is mol/L

<h3>Formula for calculating molarity :</h3>

Molarity = number of moles of solute divided by the volume of the solution\

<h3>Parameters </h3>

Molarity = ?

Number of moles is calculated using :

Mass of the solute = 3.11 grams of NaOH

Molar mass of the solute = Na + O + H

Let's input the atomic numbers of the elements: Na (23) , O (16) , H (1)

Molar mass of NaOH = 23 + 16 + 1 = 40 g/ mol

Number of moles = 3.11 ÷ 40 = 0.077 mol

Volume of the solution is measured in Liters

Let's convert 300 milliters to liters = 300 divided by 1000 = 0.3 Liters

Molarity = 0.077 ÷ 0.3 = 0. 256 mol/ L

Therefore, the Molarity of the solution is 0. 256 mol/ L

Read more on molarity here :

brainly.com/question/26873446

3 0

1 year ago

If the valency of oxygen is 2, derive the valencies of the other elements present in the following oxides. Na2O, ZnO , Al2O3 , M

11Alexandr11 [23.1K]

Answer:

1)Na2O

let the valency of Na is x

2(x)+(2)=0

2x+2=0

2x=-2

x=-1

2)ZnO

let the valency of Zn is x

x+2=0

x=-2

3)Al2O3

let the valency of Al is x

2(x)+3(2)=0

2x+6=0

2x=-6

x=-3

4)MgO

let the valency of Mg is x

x+2=0

x=-2

7 0

3 years ago

Read 2 more answers

Although both N2 and 02 are naturally present in the air we breathe, high levels of NO and NO2 in the atmosphere occur mainly in

joja [24]

Answer:

(a) Increasing the temperature adds heat which is a reactant shifting the equilibrium rightwards.

(b) Pressure has no effect since the change in the number of moles is zero.

Explanation:

Hello,

In this case, one could represent the given reaction as:

$N_2(g)+O_2(g)+ \Delta _rH \leftrightarrow 2NO$

Since it is endothermic. Thus, solving the (a) statement, one identifies the heat as a reagent, that is why the reaction cools down as it progress, therefore, by increasing the temperature, heat is added, that is, a reagent is added, which shifts the equilibrium rightwards, in other words, more NO is produced so its concentration increases.

Furthermore, for the (b) statement, since the change in the number of moles is zero, based on the stoichiometric coefficients as shown below:

$\Delta \nu =2-1-1=0$

Such value implies that the pressure has no effect on the concentration, taking into account the following form of the law of mass action:

$Kp=Kc(RT)^{\Delta \nu }$

Thus, since $\Delta \nu =0$ , $Kp=Kc$ , so no effect in concentration is due to the pressure.

Best regards.

6 0

3 years ago