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1 year ago

What is the mathematical relationship between the independent and dependent variables

Chemistry

1 answer:

Lady bird [3.3K]1 year ago

7 0

Answer:

The independent variable influences change in the dependent variable whether the points on the scatterplot go up or down from left to right.

Explanation:

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V.T2 = V2T, is an expression of

VashaNatasha [74]

Answer:

It corresponds to Charles' Law

Explanation:

Charles's Law corresponds to one of the gas laws, where temperature and volume are related, to constant pressure. That is, according to said equation, the volume of a gas varies directly with the temperature, under conditions of constant pressure.

3 0

2 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

A(n) _____ is equivalent to an electron. gamma ray omega particle alpha particle beta particle

ladessa [460]

Answer:

beta particle

Explanation:

Alpha beta and gamma radiations are the examples of ionizing radiations. When an atom is an excited state and having high energy, the atom is in unstable state. The excess of energy is released by the atom to get the stability. The released energy is in the form of radiations which may include alpha, beta, gamma, X-ray etc.

¹⁴₆C → ¹⁴₇N + ⁰₋₁e

The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.

Beta radiations:

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiations

3 0

3 years ago

How to do q solution, qrxn, moles of Mg , and delta Hrxn?

Helga [31]

Answer:

14, 508J/K

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

m = mass of substance (g) = 0.1184g

1 mole of Mg - 24g

n moles - 0.1184g

n = 0.0049 moles.

Also, q = m × c × ΔT

Heat Capacity, C of MgCl2 = 71.09 J/(mol K)

∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)

= 14, 508 J/K/kg

ΔT= (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg

= 1,7117.7472 J °C-1 g-1

∴ ΔHrxn = q/n

=1,7117.7472 ÷ 0.1184

= 14, 508J/K

6 0

3 years ago

Periodic table, chemistry questions

ehidna [41]

Answer:

Explanation:

The Lanthanides were first discovered in 1787 when a unusual black mineral was found in Ytterby, Sweden. This mineral, now known as Gadolinite, was later separated into the various Lanthanide elements. In 1794, Professor Gadolin obtained yttria, an impure form of yttrium oxide, from the mineral.

7 0

3 years ago