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ANTONII
2 months ago
8

vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu

ndance of 99.75% .
Chemistry
1 answer:
VMariaS [2.9K]2 months ago
5 0

Explanation:

The average atomic mass of vanadium = 50.9415 amu

Isotope I of vanadium has an abundance of 99.75% = 0.9975

The atomic weight of Isotope I of vanadium, m = 50.9440 amu

Abundance of Isotope II of vanadium = (100% - 99.75%) = 0.25% = 0.0025

Atomic mass of Isotope II of vanadium, m' =?

Average atomic mass of vanadium can be expressed as:

m multiplied by the abundance of Isotope I + m' multiplied by the abundance of Isotope II

So, 50.9415 amu = 50.9440 amu × 0.9975 + m' × 0.0025

Solving for m' gives us 49.944 amu.

The atomic mass of Isotope II of vanadium is 49.944 amu.

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Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7
KiRa [2933]

The Δ H value for butane (g) is -124.7 kJ/mol.

The Δ H value for CO2 (g) is -393.5 kJ/mol.

The Δ H value for H2O (g) is -241.8 kJ/mol.

The mass of butane is 8.30 grams.

Butane has a molar mass of 58 g/mol.

Considering the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

To determine the Δ H° of the reaction:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

By substituting values, we find that

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, we will calculate how many moles of butane are in 8.30 grams.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Therefore, the total energy released during the reaction is given by,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Thus, the total heat released in the reaction is 380.14 kJ.

6 0
2 months ago
The atomic mass of 13C is 13.003355. Multiply the atomic mass of 13C by its abundance. Report the number to 8 significant digits
castortr0y [3046]

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.

4 0
1 month ago
An unknown element is found to have three naturally occurring isotopes with atomic masses of 35.9675 (0.337%), 37.9627 (0.063%)
Tems11 [2777]

Answer:

The correct choice for your inquiry is option A, Argon.

Explanation:

Isotope               Atomic mass                      Percent (%)

    1                       35.9675                              0.337

    2                      37.9627                              0.063

    3                      39.9624                            99.6

To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)

Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)

Average atomic mass = 0.1212 + 0.0239 + 39.8025

Average atomic mass = 39.9476

                   Theoretical  Atomic mass

a) Ar                         39.95

b) K                          39.10

c) Cl                         35.45

d) Ca                       40.08

                 

5 0
2 months ago
A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
VMariaS [2998]

Answer: Yes, there is sufficient sodium carbonate available.

Explanation:

In this scenario, according to the specified reaction:

Using stoichiometry, one can figure out the grams of sodium carbonate required to neutralize 1,665 g of sulfuric acid as outlined below:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)

Hence, the amount on hand is 2.0 kg, which leaves 0.2 kg as surplus, therefore:

A. Yes, there is sufficient sodium carbonate available.

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3Best regards.

5 0
1 month ago
The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the
Alekssandra [3086]

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
3 months ago
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