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matrenka [14]
3 years ago
8

C-12 and C-13 are naturally-occurring isotopes of the element carbon. C-12 occurs 98.88% of the time and C-13 occurs 1.108% of t

he time. What calculation should be used to determine the atomic mass of this element?

Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
6 0

Answer:

c) (12×0.9889) + (13×0.01108)

Explanation:

Given data:

Percentage of C-12 = 98.89%

Percentage of C-13 = 1.108%

Atomic mass = ?

Solution:

98.89/100 = 0.9889

1.108/ 100 = 0.01108

Atomic mass = (12×0.9889) + (13×0.01108)

Atomic mass = (11.8668 + 0.144034)

Atomic mass = 12.01084

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Suppose you were to make a ring from a single strand of gold atoms. How many gold atoms would be required to make such a ring? H
Rudiy27
Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:

54.3 mm = 5.43 cm

Now, we determine the number of gold atoms that will be present in this:

5.43 / 1 x 10⁻⁹

There will be 5.43 x 10⁹ atoms


We now determine the number of moles this is by:

one mole = 6.02 x 10²³ atoms

Moles = 5.43 x 10⁹ / 6.02 x 10²³ 
Moles = 9.01 x 10⁻¹⁵ moles

The molar mass of gold is 197 g/mol

The mass is 9.01 x 10⁻¹⁵  * 197

The mass of the strand is 1.76 x 10⁻¹² grams
8 0
3 years ago
Read 2 more answers
A chemical reaction produced 177.3 grams of chlorine has (Cl²). What was the volume of the gas?
vichka [17]
Since there is so little information given, I will assume that we are at STP and i can use the conversion factor at STP--->> 22.4 Liters= 1 mol of gas

before we use this conversion, we need to convert the grams to moles using the molar mass of the molecule.

molar mass of Cl₂= 35.5 x 2= 71.0 g/ mol

177.3 g (1 mol/ 71.0 g)= 2.50 mol Cl₂

then we use the conversion to get the volume

2.50 mol Cl₂ (22.4 Liters/ 1 mol)= 55.9 Liters
3 0
3 years ago
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
How is a sodium ion symbol written? so so- na na-
vfiekz [6]
A sodium ion symbol is written as Na^+.
6 0
3 years ago
Read 2 more answers
What is the redox half equation for 3Ag2S + 2Al --&gt; 6Ag + Al2, and identity which material is oxidized and which is reduced?
marta [7]

Answer:

Al is oxidized while Ag is reduced.

Explanation:

The complete molecular equation is;

3Ag2S + 2Al --> 6Ag + Al2S3

Oxidation half equation;

2Al ------> 2Al^3+ + 6e

Reduction half equation;

6Ag^+ + 6e -------> 6Ag

Overall redox reaction equation;

2Al + 6Ag^+ ----->2Al^3+ + 6Ag

Hence; Al is oxidized while Ag is reduced.

5 0
3 years ago
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