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3 years ago

Question 10 of 23

Chemistry

1 answer:

Margarita [4]3 years ago

7 0

Answer: O D. A digital signal moves between a discrete number of values.

Explanation:

The digital signals are considered more reliable over the analog signals as they encode the information in a coded form. The bits or samples of the data are transmitted and converted into digital and numerical value. The stream of encoded data is in the form of continuous data at regular time intervals. It provides information in waveform and the data is in compact form. The data is in the form of binary bits 0 and 1 so greater the number of bits greater will be the greater will be the resolution of the information.

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CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s

devlian [24]

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The combustion of $CO$ will be,

$CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$ $\Delta H_1=-110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mol$

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) $CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)$ $\Delta H_1=110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mol$

The expression for enthalpy change for the reaction will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2$

$\Delta H_{rxn}=(110.5)+(-393.5)$

$\Delta H_{rxn}=-283kJ/mol$

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

6 0

3 years ago

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 53.88 g of MgI₂ form, what is the pe

ad-work [718]

We know the law of conservation of mass

It states that mass is neither formed nor destroyed in any chemical reaction.
Mass of reactants=Mass of products.

Here

Mg and I_2 are reactants
MgI_2 is product with some yield.
Mass of reactants=10+60.0=70.0g
Mass of MgI_2=53.88g
Mass of yield=Product-MgI_2=70-53.88=16.12g

Lets find the percentage

$\\ \tt\hookrightarrow \dfrac{Mass\:of\:yield}{Total\:mass}\times 100$

$\\ \tt\hookrightarrow \dfrac{16.12}{70}\times 100$

$\\ \tt\hookrightarrow 0.23028(010)=23.028\%$

6 0

2 years ago

Read 2 more answers

Work file: climate and weather; what’s the difference

Deffense [45]

Answer: Weather refers to short term atmospheric conditions while climate is the weather of a specific region averaged over a long period of time. Climate change refers to long-term changes.

7 0

3 years ago

Meta-bromoaniline was treated with nano2 and hcl to yield a diazonium salt. Draw the product obtained when that diazonium salt i

Y_Kistochka [10]

Answer:

I believe is the correct answer

Explanation:

100bromoaniline+nano2+hcl

4 0

2 years ago

Using the molecular orbital model to describe the bond- ing in F2????, F2, and F2????, predict the bond orders and the relative

masya89 [10]

Answer: F2 : bond order= 1.0

F2+: bond order = 1.5

F2- : bond order = 0.5

Explanation:

1. Starting with F2+

The configuration gives;

F2+ = 9F = 1S2.2S2.2P5

= 9F+ = 1S2.2S2.2P4 (this shows it gives out an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py1

The number of Electrons = (9*2) – 1 = 18 -1 = 17

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 7

Bond order = (10-7)/2 = 3/2 = 1.5

Number of unpaired electrons = 1

2. Starting with F2

The configuration gives;

F2 = 9F = 1S2.2S2.2P5

9F = 1S2.2S2.2P5 (this shows no loss of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2

The number of Electrons = (9*2) = 18 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 8

Bond order = (10-8)/2 = 2/2 = 1.0

Number of unpaired electrons = 0

3. Starting with F2-

The configuration gives;

F2- = 9F = 1S2.2S2.2P5

10F--= 1S2.2S2.2P6 (this shows an addition of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2 σ*2Pz

The number of Electrons = (9*2) + 1 = 19 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 9

Bond order = (10-9)/2 = 1/2 = 0.5

Number of unpaired electrons = 1

To get the order of bond as well as length, we know that;

Bond order directly proportional to 1/ Bond length

Therefore the Ascending Bond length = F2+ ˂ F2 ˂ F2-

3 0

3 years ago