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Kamila
9 days ago
5

The atomic mass of 13C is 13.003355. Multiply the atomic mass of 13C by its abundance. Report the number to 8 significant digits

.
Chemistry
1 answer:
castortr0y [3K]9 days ago
4 0

The result is 0.14303691.

Carbon-13 (¹³C) is a stable isotope of carbon with a mass number of 13, composed of six protons and seven neutrons.

Isotopes are elements that share the same atomic number but have different mass numbers, meaning they have a varying number of neutrons.

ω(¹³C) = 1.10% ÷ 100%.

ω(¹³C) = 0.0110; this indicates the natural abundance of carbon-13.

m(¹³C) = 13.003355; the atomic mass assigned to carbon-13.

ω(¹³C) · m(¹³C) = 0.0110 · 13.003355.

ω(¹³C) · m(¹³C) = 0.14303691.

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A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empi
Tems11 [2777]
Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO
3 0
1 month ago
How many protons neutrons and electrons are there in a neutral atom of 43k (potassium-43)?
castortr0y [3046]

Answer:

             Protons: 19

             Neutrons: 25

             Electrons: 19

Explanation:

Protons:

                   The atomic number determines the number of protons in an atom. Consequently, with Potassium's atomic number being 19, it contains 19 protons.

Neutrons:

                   The formula to find neutrons is:

                           # of Neutrons  =  Atomic Mass - # of Protons

Given:

                                Atomic Mass  =  43

                                # of Protons  =  19

Thus,

                                # of Neutrons  =  43 - 19

                                # of Neutrons  =  24

Electrons:

                                   In a neutral atom, the quantity of electrons matches that of protons. Therefore, a neutral Potassium atom with 19 protons must equally have 19 electrons.

3 0
1 month ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
KiRa [2933]

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

\frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

\frac{0.6}{298.2}   = \frac{P_{2} }{345.4}

P₂ = 0.7atm

3 0
1 month ago
A sample of a monoprotic acid (ha) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous naoh. if 30.
Alekssandra [3086]
The formula for a monoprotic acid can be represented as HA, and its reaction with a base is shown as follows: HA + NaOH ---> NaA + H₂O. The stoichiometry between the acid and the base is 1:1. At the point of neutralization, the moles of HA equals the moles of the base. The moles of NaOH that reacted can be calculated as 0.100M / 1000 mL/L x 30.0 mL = 0.003 mol. Consequently, the moles of HA that reacted equal 0.003 mol. The mass of the acid is 0.384 g, yielding a molar mass of 0.384 g / 0.003 mol = 128 g/mol.
3 0
1 month ago
In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
KiRa [2933]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

\Delta H =\frac{Q}{n}

To calculate the heat (Q):

Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J

Next, we need to calculate the number of moles involved in the reaction (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

With these two values, we can substitute them into the first equation:

\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole

Have a great day!

5 0
1 month ago
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