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3 years ago

What order do these go in? please help

Chemistry

2 answers:

pochemuha3 years ago

5 0

Answer:

Chemical energy found in the bonds between atoms

Nuclear energy found in nucleus of an atom

Electrical energy found in the flow of electrical charges

Thermal energy found when internal energy is transferred.

Jlenok [28]3 years ago

3 0

Answer:

chemical energy - found in the bonds between atoms

nuclear energy - found in the nucleus of an atom

electrical energy - found in the flow of electric charges

thermal energy - found when internal energy is transferred

Explanation:

hope this helps

plz mar brainliest

92% sure its correct

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Describe how visible spectroscopy could be used to determine the order of a reaction in which a single colorless reactant produc

d1i1m1o1n [39]

Answer:

Explanation:

In a reaction, where, one of the reactant produces a colored product, visible spectroscopy can be used to determined the order of a reaction, the change in concentration of the reactant which forms the colored product is determined by absorbance measurement over time. The data for the concentration and time are plotted on the y and x axis and If we get a straight line it is a zero-order reaction. If instead, a plot of ln[concentration] versus time gives a straight line, it is a first order reaction. However, If 1/concentration versus time gives a straight line, it is a second order reaction kinetics. The other reactants may be changed while keeping this reactant as constant and change on rate of the reaction is observed to see If the other reactant affects the reaction or not.

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4 years ago

Show the relationship of number of moles and volume in Avogadro's law using data. Write your calculations and answer on a separa

leonid [27]

Answer:

See below ~

Explanation:

The calculated values of V/n :

⇒ 1.5/0.3 = 5

⇒ 3/0.6 = 5

⇒ 4.5/0.9 = 5

⇒ 6/1.2 = 5

⇒ 7.5/1.5 = 5

1. From this we understand that the calculated values of V/n remain constant, equal to <u>5</u> in this case.

2. The volume-mole graph will be a straight line passing through the origin. (Attached below)

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2 years ago

For IR radiation with û = 1,130 cm 1, v=__THz

Dmitry [639]

<u>Answer:</u> The frequency of the radiation is 33.9 THz

<u>Explanation:</u>

We are given:

Wave number of the radiation, $\bar{\nu}=1130cm^{-1}$

Wave number is defined as the number of wavelengths per unit length.

Mathematically,

$\bar{\nu}=\frac{1}{\lambda}$

where,

$\bar{\nu}$ = wave number = $1130cm^{-1}$

$\lambda$ = wavelength of the radiation = ?

Putting values in above equation, we get:

$1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm$

Converting this into meters, we use the conversion factor:

1 m = 100 cm

So, $8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m$

The relation between frequency and wavelength is given as:

$\nu=\frac{c}{\lambda}$

where,

c = the speed of light = $3\times 10^8m/s$

$\nu$ = frequency of the radiation = ?

Putting values in above equation, we get:

$\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}$

$\nu=0.339\times 10^{14}Hz$

Converting this into tera Hertz, we use the conversion factor:

$1THz=1\times 10^{12}Hz$

So, $0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz$

Hence, the frequency of the radiation is 33.9 THz

7 0

3 years ago

A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is conve

Helen [10]

Answer:

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

Explanation:

The differential equation is given as:

$\frac{dC}{dt} = r- kC$

$\frac{dC}{r- kC} = dt$

Taking integral of both sides; we have:

$\int\limits \frac{dC}{r- kC} = \int\limits dt$

$-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD$

$r-kC=e^{-kt-kD}$

$r-kC=e^{-kt}e^{-kD}$

$r-kC=Ae^{-kt}$

$kC=r-Ae^{-kt}$

$C=\frac{r}{k}-\frac{A}{k}e^{-kt}$

$C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}$ ------- equation (1)

If C(0)= $C_o$ ; we have:

C(0)= $\frac{r}{k}-\frac{A}{k}e^0$ (where; A is an integration constant)

$C_o = \frac{r}{k}- \frac{A}{k}$

$C_o=\frac{r-A}{k}$

$kC_o=r-A$

$A=r-kC_o$

Substituting $A=r-kC_o$ into equation (1); we have;

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

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3 years ago

What is the general formula for a secondary amine? rnh2 h2nrnh2 rn2h r2nh

zvonat [6]

The general fomula of the amine group ought to be R2NH because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

<h3>What is a secondary amine?</h3>

A secondary amine is one in which two alkyl groups are subtituted on the nitrohen atom of ammonia.

This implies that the general fomula of the amine group ought to be R2NH. This is because, the two hydrogen atoms of ammonia have been substituted by alkyl groups.

Learn more about secondary amine: brainly.com/question/12682525

4 0

2 years ago