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1 year ago

Lithium is located in the first group of the periodic table. How many valence electrons does the element lithium have?

Chemistry

1 answer:

e-lub [12.9K]1 year ago

6 0

Answer: 1

Explanation: Lithium has a single electron in the second principal energy level, and so we say that lithium has one valence electron.

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Mg +2+ 2e- → Mg oxidation or reduction

ahrayia [7]

Answer:

Reduction

Explanation:

The charge on Mg goes from 2+ to 0 which means this is a reduction.

6 0

3 years ago

Convert 1.422g/cm^2 to mg/mm^2

Natalija [7]

The answer is 14.22 mg / (mm^2)

6 0

2 years ago

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

4 0

3 years ago

For the reaction of hydrogen with iodine

fenix001 [56]

Answer:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Explanation:

Hello!

In this case, considering the given chemical reaction:

$H_2(g) + I_2(g) \rightarrow 2HI(g)$

Thus, by applying the law of rate proportions, we can write:

$\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}$

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Best regards!

8 0

2 years ago

What are some of the body's internal defenses

Amanda [17]

What are some of the body's internal defenses. The skin helps protect stuff to!

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2 years ago

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