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3 years ago

What electrically neutral atom will have an electron configuration of 1 s 22 s 22 p 4? Use a periodic table.

Chemistry

1 answer:

a_sh-v [17]3 years ago

7 0

Answer:

Explanation:

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From the following data, calculate the average bond enthalpy for the NOH bond: NH3(g) ¡NH2(g) 1 H(g) ¢H° 5 435 kJ/mol NH2(g) ¡NH

Sphinxa [80]

Answer:

The average bond enthalpy for the N-H bond is 392 kJ/mol.

Explanation:

$NH_3(g) \rightarrow NH_2(g) +1 H(g),\Delta H^o_{1}= 435 kJ/mol$ ..[1]

$NH_2(g) \rightarrow NH(g) + H(g),\Delta H^o_{2}= 381 kJ/mol$ ..[2]

$NH(g) \rightarrow N(g)+ 1 H(g),\Delta H^o_{3}= 360 kJ/mol$ ..[3]

On adding [1] , [2] and [3] , we get:

$NH_3(g) \rightarrow N(g) +3H(g),\Delta H^o_{4}= ?$

$\Delta H^o_{4}=\Delta H^o_{1}+\Delta H^o_{2}+\Delta H^o_{3}$

$= 435 kJ/mol+381 kJ/mol +360 kJ/mol = 1,176 kJ/mol$

There three bonds N-H bond sin ammonia. Then average bond enthalpy of single N-H bond is :

$\Delta H^o_{N-H}=\frac{1,176 kJ/mol}{3}=392 kJ/mol$

The average bond enthalpy for the N-H bond is 392 kJ/mol.

4 0

3 years ago

5.4×10^3 × 1.2×10^7

Nadya [2.5K]

5.4*10^3 * 1.2*10^7.

5.4 * 1000 * 1.2 * 10000000

Multiply them = 64800000000

4 0

3 years ago

If you add a large piece of zinc to a beaker of acid and, in a separate beaker, zinc shavings to a beaker of acid (of equal conc

Sergeu [11.5K]

Answer: surface area

Explanation:

Surface area affects rate of reaction. The smaller the surface area the faster the rate of reaction. The zinc shavings have a smaller surface area of reactants compared to the large piece of zinc. The smaller surfaces area of reactants ensures that it comes into close proximity with the acid solution for reaction to take place.

7 0

3 years ago

Help please!!!!!

weqwewe [10]

Answer:

OPTION B..

Explanation:

A major environmental concern related to nuclear power is the creation of radioactive wastes such as uranium mill tailings, spent (used) reactor fuel, and other radioactive wastes. These materials can remain radioactive and dangerous to human health for thousands of years..

6 0

3 years ago

A sample of Al2O3 decomposed to pure aluminum and oxygen. After decomposition, 29.0 moles of Al were recovered. How many moles o

Rina8888 [55]

The decomposition of Al2O3 to form pure aluminum (Al) along with the release of oxygen (O2) gas can be represented by the following chemical reaction:

2Al2O3 → 4Al + 3O2

Based on the reaction stoichiometry:

2 moles of aluminum oxide decomposes to form 4 moles of pure aluminum

Therefore, if 29.0 moles of aluminum is recovered, then the moles of aluminum oxide required will be:

= 29.0 moles Al * 2 moles Al2O3/4 moles Al

= 14.5 moles of Al2O3

4 0

4 years ago