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mariarad [96]
3 years ago
15

The feeling of which orbital is represented by the transition metals in period Four

Chemistry
1 answer:
Fudgin [204]3 years ago
4 0
The best answer is Titanium
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Earth's axis is tilted. Tell what would happen if it was vertical instead.
slamgirl [31]
I think the like term is 6 hope this is right
7 0
3 years ago
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how many molecule of carbon dioxide are needed to react with excess iron oxide to produce 11.6 g of iron
lesya692 [45]

Answer:

0.16 moles of Carbon

Explanation:

The balanced reaction equation:

2Fe_{2}O_{3} + 3C → 4Fe + 3CO_{2}↑

The mole ratio of Carbon to Iron is 3 : 4 (since Fe2O3 is in excess)

i.e 3 moles of C produces 4 moles of Fe.

If 1 mole of Fe - 55.8g of Fe

? moles - 11.6g of Fe

= \frac{11.6}{55.8} = 0.208 moles

But 3 moles of C - 4 moles of Fe

? moles of C - 0.208 moles of Fe

= \frac{3 *0.208}{4} = 0.16 moles of carbon.

I hope this explanation was clear and useful.

5 0
3 years ago
The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
a mineral that has - can be split fairly easily along plains with weak atomic attraction A. streak B. Cleavage C. luster D. Hard
tatyana61 [14]
I am going to have to say C. luster if not then B. Cleavage
5 0
3 years ago
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A hypothetical element x has 3 naturally occurring isotopes: 41.20% of 21x, with an atomic weight of 21.016 amu, 50.12% of 22x,
iogann1982 [59]
The average atomic weight is calculated by adding up the products of the percentage abundance and atomic weight. In this item, we have the equation,

     A = (0.412)(21.016 amu) + (0.5012)(21.942 amu) + (0.0868)(23.974 amu)

Simplifying the operation will give us the answer of 21.74 amu.

<em>Answer: 21.74 amu</em>
4 0
3 years ago
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