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vovangra
5 days ago
9

in sample of elemental bromine, 55% or the atoms are Br-79, and the remainder are Br-81. if this sample is typical of naturally

occurring bromine, what is the average atomic mass of bromine?
Chemistry
1 answer:
KiRa [971]5 days ago
3 0

Answer:

The typical atomic weight of bromine is 79.9 amu.

Explanation:

Provided information:

Br⁷⁹ constitutes 55% of the sample

Br⁸¹ constitutes 45% of the sample

What is the average atomic weight of bromine?

Calculation formula:

Average atomic mass = [isotope mass × its proportion] + [isotope mass × its proportion] +...[ ] / 100

We can now insert the known values into the formula.

Average atomic mass = [55 × 79] + [81 × 45] / 100

Average atomic mass = 4345 + 3645 / 100

Average atomic mass = 7990 / 100

Average atomic mass = 79.9 amu

The typical atomic weight of bromine is 79.9 amu.

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Which of the following reactions is a redox reaction? (a) K2CrO4 + BaCl2 → BaCrO4 + 2KCl (b) Pb2+ + 2Br- → PbBr2 (c) Cu + S → Cu
castortr0y [923]

(c) Cu + S → CuS is classified as a redox reaction

Explanation:

The following reactions are presented:

(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl

(b) Pb²⁺ + 2 Br⁻ → PbBr₂

(c) Cu + S → CuS

Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:

Cu + S → CuS

In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.

Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).

Learn more about:

redox reactions

7 0
1 day ago
The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the
Alekssandra [968]

Answer:

Mitochondria are plentiful in mammalian cells, with their proportions varying across different tissues, from less than 1% in white blood cells to as high as 35% in heart muscle cells. It is essential to understand that mitochondria are not static structures but instead form a dynamic network that frequently undergoes processes of fission and fusion. In skeletal muscle, they exist as part of a reticular membrane network. The two subpopulations, subsarcolemmal (SS) and intermyofibrillar (IMF) mitochondria, occupy different subcellular regions and exhibit slight differences in their biochemical and functional characteristics tied to their anatomical context. The SS mitochondria are positioned just beneath the sarcolemma, while IMF mitochondria are found closely associated with myofibrils. Their distinct properties likely play a role in their adaptability. SS mitochondria make up about 10-15% of the total mitochondrial volume and are believed to be more adaptable than their IMF counterparts, despite the latter displaying higher levels of protein synthesis, enzyme activity, and respiration (1).

Explanation:

0 0
16 days ago
Read 2 more answers
he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp
eduard [944]

1) Drift velocity: 3.32\cdot 10^{-4}m/s

2. 5.6\cdot 10^{13} electrons per individual

Explanation:

1)

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

v_d=\frac{I}{neA}

where

I stands for current

n represents the density of free electrons

e=1.6\cdot 10^{-19}C indicates the charge of an electron

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

A=\pi r^2

where r denotes the wire's radius. Thus, the equation transforms to

v_d=\frac{I}{ne\pi r^2}

In this scenario, we have:

I = 8.0 A as the current

8.5\cdot 10^{28} m^{-3} indicates the free electron concentration

d = 1.5 mm is the diameter, making the radius

r = 1.5/2 = 0.75 mm = 0.75\cdot 10^{-3}m

So, the resulting drift velocity is:

v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s

2)

The entire length of the cord is

L = 3.00 m

And the cross-sectional area is

A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2

Consequently, the volume of the cord is

V=AL (1)

The number of electrons per unit volume is n, thus the total electrons in this cord would be

N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}

Overall, the Earth's population rounds to 8 billion individuals, equating to

N'=8\cdot 10^9

Hence, the number of electrons distributed to each person is:

N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}

7 0
5 days ago
Which of the following statements is true about the relationships between photon energy, wavelength, and frequency?
KiRa [971]

Answer: The Answer is A.

Explanation:

The energy of a photon is directly related to its electromagnetic frequency, meaning it is inversely related to the wavelength. A higher frequency results in greater energy for the photon. Conversely, a longer wavelength corresponds to lower energy levels.

Hope this Helps!

8 0
3 days 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 [971]

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
5 days ago
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