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3 years ago

How is a lithium atom (Li) different from a lithium ion (Li+)?

Chemistry

2 answers:

timofeeve [1]3 years ago

8 0

Answer:

Li has fewer protons than Li+

zalisa [80]3 years ago

8 0

Li has fewer electrons than Li+

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Use the atomic mass of indium to calculate the relative abundance of indium-113.

ASHA 777 [7]

The relative abundance of indium-113 is 4%.

The isotopes are species of the same element having the same atomic number but a different mass number.

The elements occurring in nature exist as multiple isotopes.

When we take into account the existence of these isotopes and their relative abundance (percent), the average atomic mass of that element can be computed, which is given by the following formula,

Average atomic Mass= (%age of isotope 1) x (Mass of isotope 1) + (%age of isotope 2) x (Mass of isotope 2)/100

Indium exists in the form of Indium-113 and Indium-115.

The mass of Indium-113 is 112.90 u.

The mass of Indium-115 is 114.90 u.

The average atomic mass of Indium is 114.82 u.

Let the %age of isotope 1(Indium-113) be X.

Then, the %age of isotope 2(Indium-115) would be 100-X.

Applying the values in the formula,

Average atomic mass = 112.90X+114.90(100-X)

114.82 = 112.90X+114.90(100-X)

On solving the above equation, the value of X comes out to be 4%.

Thus, the relative abundance/%age abundance of Indium-113 is 4%.

To know more about "Average Atomic Mass", refer to the following link:

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7 0

1 year ago

How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?

ankoles [38]

The chemical reaction equation for this is

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6,

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x $\frac{1 mole XeF6}{3 moles H2}$ x $\frac{245.28 g XeF6}{1 mole XeF6}$ = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.

3 0

3 years ago

you have been observing an insect that defends itself from enemies by secreting a caustic liquid. analysis of the liquid shows i

Alexus [3.1K]

pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions

[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate

[HCOOH] = 1.45 M - 0.015 M

[HCOOH] = 1.435 M; equilibrium concentration of formic acid

pKa = -logKa

pKa = -log 1.8×10⁻⁴ M

pKa = 3.74

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 3.74 + log (0.015 M/1.435 M)

pH = 3.74 - 1.98

pH = 1.76

More about buffer: brainly.com/question/4177791

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4 0

1 year ago

Blood samples for research or medical tests sometimes have heparin added. Why is this done?

alina1380 [7]

It is essential for accurate results that the correct volume of blood is sampled to achieve a correct concentration (and dilution, if liquid heparin is used), and that blood and anticoagulant are well mixed immediately after sampling.

8 0

3 years ago

Given 7.40 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

Dafna11 [192]

On complete conversion (100% yield) 9.75 g of ethyl butyrate will be produced. Below is the solution.....

4 0

3 years ago