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Dima020 [189]
3 years ago
5

The atomic weight of a newly discovered element is 98.225 amu. it has two naturally occuring isotopes. one isotope has a mass of

96.780 amu. the second isotope has a percent abundance of 41.7%. what is the mass of the second isotope?
Chemistry
1 answer:
Aloiza [94]3 years ago
4 0
Atomic weight of an element can be calculated as follows:
average atomic weight = 
(atomic weight of first isotope)(its percentage of abundance) + 
(atomic weight of second isotope)(its percentage of abundance)

average atomic weight = 98.225 amu
atomic weight of first isotope = 97.780 amu
its percentage of abundance = 1 - 0.417 = 0.583
atomic weight of second isotope = ??
its percentage of abundance = 0.417

So, just substitute in the above equation to get the atomic weight of the second isotope as follows:
98.225 = (97.78)(0.583) + (mass2)(0.417) 
atomic weight of second isotope = 98.847 amu
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Determina el pH del café cuya concentración de H+ es de 0.00001M
lara31 [8.8K]

Answer:

5

Explanation:

Given parameters:

Hydrogen ion concentration  =  0.00001M

Unknown:

pH of the solution =?

Solution:

The pH is used to estimate the degree of acidity or alkalinity of a solution. To solve for pH of any solution, we use the expression below;

          pH  = -log [H⁺]

[H⁺] is the hydrogen ion concentration

        pH  = -log (1 x 10⁻⁵)

      pH = -(-5) = 5

4 0
3 years ago
I could really use some help with this question, thank you in advance!​
Taya2010 [7]
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Explanation Plato
6 0
2 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
How much (Q) heat is needed to melt 35 g of iodine? Hf = 61.7 J/g.
aev [14]

Taking into account the definition of calorimetry and latent heat, a heat of 2159.5 J is needed to melt 35 g of iodine.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

<h3>Heat needed to melt iodine</h3>

In this case, you know:

  • m= 35 g
  • L=61.7 \frac{J}{g}

Replacing in the definition of latent heat:

Q= 35 g× 61.7 \frac{J}{g}

Solving:

<u><em>Q=2159.5 J</em></u>

Finally, a heat of 2159.5 J is needed to melt 35 g of iodine.

Learn more about calorimetry:

brainly.com/question/14057615

brainly.com/question/24988785

brainly.com/question/21315372

brainly.com/question/13959344

brainly.com/question/14309811

brainly.com/question/23578297

#SPJ1

8 0
2 years ago
A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the
yarga [219]

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X  1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of  H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

5 0
1 year ago
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