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VMariaS [17]
3 years ago
7

If you have 100.0 grams of carbon and 100.0 grams or iron, which sample will contain more atoms

Chemistry
1 answer:
vodomira [7]3 years ago
6 0
First you need to solve for the number of moles in each. 
1 mole of anything = atomic mas of that element (g)

Carbon has an atomic mass of 12.0107
1 mole of Carbon = 12.0107 g

Iron has an atomic mass of 55.845
1 mole of Iron = 55.845 g

now convert each from grams to moles 
(grams of Carbon cancel and you are left with moles of carbon)
<u>100.0 g Carbon x 1 mol Carbon </u> = 8.326 mol Carbon
       12.0107 g Carbon

(grams of Iron will cancel and you are left with moles of iron)
<u>100.0 g Iron x 1 mol Iron</u> = 1.791 mol Iron
      55.845 g Iron

now we can use avogadro's number to solve for the amount of atoms in each 
1 mol = 6.02x10^23 atoms

<u>8.326 mol Carbonx 6.02 x 10^23 atoms </u> = 5.012 x 10^24 atoms Carbon<u>
</u>        1 mol Carbon
<u>
</u><u>1.791 mol Iron x 6.02 x 10^23 atoms </u> = 1.078 x 10^24 atoms Iron<u>
</u>        1 mol Iron
<u>
</u>5.012 x 10^24 atoms Carbon > 1.078 x 10^24 atoms Iron
<u>
</u>
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HELP ASAP!!!
givi [52]
This law states that, despite chemical reactions or physical transformations, mass is conserved — that is, it cannot be created or destroyed — within an isolated system. In other words, in a chemical reaction, the mass of the products will always be equal to the mass of the reactants.



I looked this up but hope it helps!
4 0
3 years ago
24.3 2 An artifact classified as seeds, found in a site at Newlands Cross, Ireland, is found to have a 14C radioactivity of 0.10
Rina8888 [55]

Answer:

Age ≅ 7500 years

Explanation:

All radioactive decay is 1st order kinetics and described by the expression

A = A₀e^-kt => t = ln(A/A₀) / -k

k = 0.693 / t(half life) = (0.693 / 5730)yrs⁻¹ = 1.21 x 10⁻⁴ yrs⁻¹

t = Age = [ln(0.103/0.255) / - 1.21 x 10⁻⁴] yrs = 7500 years

5 0
3 years ago
Match each of the following topical dosage forms with its correct definition.
Dmitry [639]

Answer:

cream - contains a higher proportion of oil than water

ointment - dr4g mixed in approximately equal proportions of oil and water

i don't know about the other two sorry

8 0
1 year ago
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
An increase in the atomic number________the atomic radius moving from left to right across a period.
frez [133]

Answer:

decrease

Explanation:

Atomic radius :

It is the distance from the center of nucleus to the outer most electronic shell.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required

4 0
3 years ago
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