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3 years ago

A 24.0 gram sample of copper was

Chemistry

1 answer:

musickatia [10]3 years ago

5 0

Answer:

3.84 J/g°C

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass of substance

c = specific heat of copper

∆T = change in temperature (°C).

Based on the provided information;

Q = 43783J

m = 24g

∆T = 500°C - 25°C = 475°C

c = ?

Using Q = m × c × ∆T

43783 = 24 × c × 475

43783 = 11400c

c = 43783 ÷ 11400

c = 3.84 J/g°C

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Use coulomb's law to calculate the ionization energy in kj/mol of an atom composed of a proton and an electron separated by 185.

Tems11 [23]

Coulomb's law mathematically is:
F = kQ₁Q₂/r²
we integrate this with respect to distance to obtain the expression for energy:
E = kQ₁Q₂/r; where k is the Coulomb's constant = 9 x 10⁹; Q are the charges, r is the seperation
Charge on proton = charge on electron = 1.6 x 10⁻¹⁹ C
E = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹) / (185 x 10⁻¹²)
E = 1.24 x 10⁻¹⁸ Joules per proton/electron pair
Number of pairs in one mole = 6.02 x 10²³
Energy = 6.02 x 10²³ x 1.24 x 10⁻¹⁸
= 746.5 kJ

5 0

3 years ago

Read 2 more answers

B. The heat of reaction for the process described in (a) can be determined by

julsineya [31]

Answer: I believe the 1st and 3rd reactions are better obtained through reference sources and the 2nd and 4th are easiest and safest to measure in the laboratory.

Explanation:

I am also working on this Pre-lab right now, and I looked back at the first question to help get my answer. In the first question (a), it is noted that ammonia gas and gaseous hydrochloric acid are both potentially dangerous in gaseous form. I saw that both the 1st and 3rd reactions contained noxious gases (I knew this because there was a (g) in both of these reactions). Using the knowledge from the first question that the noxious gases were potentially dangerous, I assumed that those reactions were the ones that are better obtained through the reference sources. The 2nd and 4th reactions did not contain any noxious gases, so I assumed those ones were easiest and safest to measure in the laboratory. Hope this helps!

7 0

4 years ago

Arrange the eluting power of the listed solvents in normal-phase chromatography. The highest eluting power solvent is number 1;

BabaBlast [244]

Answer:

Explanation:

Part a

Normal Phase Chromatography

In the normal phase chromatography, the eluting power is highest for the solvent which has the highest polarity.

So from the given solvents as Methanol, Dichloromethane and Ethyl Acetate

the relative polarities are given as

Methanol=0.762

Dichloromethane (methylene chloride)=0.309

Ethyl Acetate=0.228

So the numbers are given as

Methanol :1

Dichloromethane (methylene chloride):2

Ethyl Acetate:3

Reverse Phase Chromatography

In the reverse phase chromatography, the eluting power is highest for the solvent which has the lowest polarity.

So from the given solvents as diethyl ether , water and toluene

the relative polarities are given as

Diethyl ether=0.117

Water=1.0

Toluene=0.099

So the alphabets are given as

Toluene :A

Diethyl ether: B

Water:C

3 0

3 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

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3 years ago

Although hcl(aq) exhibits properties of an arrhenius acid, pure hcl gas and hcl dissolved in a nonpolar solvent exhibit none of

Tema [17]

HCL(g) it consists of covalently molecules which are bonded and they do not ionize.
Non-polar solvent molecules do not attract molecules of HCL which cause them to be ionized.

3 0

3 years ago