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3 years ago

6

Combining Boyle's law and Charles' law forms one presure-temperature-volume relationship and holds the mass of the gas constant.

Which equation represents the relationship of the combined gas law?

1 answer:

Oksanka [162]3 years ago

4 0

(P1V1/T1)=(P2V2/T2)

I just took the test and this should be correct

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There are two beakers that contain the same liquid substance at the same temperature. The larger beaker is 1,000ml and the small

azamat

Answer:

Answer choice B

Explanation:

Since you do not know the volume of the liquid in each beaker, the one in the smaller beaker could have more substance and therefore more thermal energy. If they had the same amount of substance, then the more voluminous one would radiate faster. However, since you do not know this, there is no way to tell. PM me if you have more questions. Hope this helps!

6 0

3 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

2

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

So

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

2 years ago

1. The emission spectrum of mercury atoms has a bright green line with wavelength

SashulF [63]

Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.

From the formula;

E = hc/λ

h = Plank's constant = $6.6 * 10^-34$ Js

c = speed of light= $3 * 10^8$

λ = wavelength = $546.1 * 10^-9$ m

E = $6.6 * 10^-34 * 3 * 10^8/546.1 * 10^-9$

E = $3.63 * 10^-19$ J

Also;

E =hf

Where;

h = Planks's constant

f = frequency of photon

f = E/h

f = $3.63 * 10^-19 J/6.6 * 10^-34$

f = $5.5 * 10^14$ Hz

Learn more: brainly.com/question/18415575

5 0

2 years ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

Elemental iodine (I 2) is a solid at room temperature. What is the major attractive force that exists among different I 2 molecu

Ulleksa [173]

Explanation:

As $I_{2}$ is a covalent compound because it is made up by the combination of two non-metal atoms. Atomic number of an iodine atom is 53 and it contains 7 valence electrons as it belongs to group 17 of the periodic table.

Therefore, sharing of electrons will take place when two iodine atoms chemically combine with each other leading to the formation of a covalent bonding.

Hence, weak forces like london dispersion forces will be present between a molecule of $I_{2}$ .

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

thus, we can conclude that london dispersion force is the major attractive force that exists among different $I_{2}$ molecules in the solid.

7 0

3 years ago