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kykrilka [37]
3 years ago
7

Help please all please

Chemistry
2 answers:
Luden [163]3 years ago
5 0

Answer:

I don’t want to download a pdf that I don’t know what it is…

Also, brainly strictly says that we can’t post questions about a test or quiz that is found in school…

Explanation:

Lilit [14]3 years ago
3 0

Answer:

wnich subject and which class pdf i will help you tell me

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Metallic bonding is...
kap26 [50]

Answer:

d an attraction between positive ions and electrons

7 0
2 years ago
When N2(g) reacts with H2(g) to form NH3(g), 92.2 kJ of energy are evolved for each mole of N2(g) that reacts. Write a balanced
Naddika [18.5K]

Answer:

N₂ + 3H₂ → 2NH₃   ΔH = - 92.2KJ

Explanation:

Let's write out the chemical equation between Nitrogen and Hydrogen to Form Ammonia.

Nitrogen + Hydrogen = Ammonia

N₂ + H₂ → NH₃

A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.

The balanced stoichiometric chemical equation is given as;

N₂ + 3H₂ → 2NH₃

92.2 kJ of energy are evolved for each mole of N2(g) that reacts. And from the equation, 1 mole of N2 reacts.

The enthalpy change, ΔH = - 92.2KJ. The negative sign is because heat is being evolved.

The balanced thermochemical equation;

N₂ + 3H₂ → 2NH₃   ΔH = - 92.2KJ

6 0
3 years ago
The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days
professor190 [17]

Answer:  

6.2 g  

Explanation:  

In a first-order decay, the formula for the amount remaining after <em>n</em> half-lives is  

N = \frac{N_{0}}{2^{n}}  

where  

<em>N</em>₀ and <em>N</em> are the initial and final amounts of the substance  

1. Calculate the <em>number of half-lives</em>.  

If t_{\frac{1}{2}} = \text{3.8 da}  

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}

2. Calculate the <em>final mass</em> of the substance.  

\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}

4 0
3 years ago
A certain process has ΔH° &gt; 0, ΔS° &lt; 0, and ΔG° &gt; 0. The values of ΔH° and ΔS° do not depend on the temperature. Which
tino4ka555 [31]

Answer: It is non-spontaneous at all T.

Explanation:

According to Gibb's equation:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy = +ve

\Delta H = enthalpy change  = +ve

\Delta S = entropy change  = -ve

T = temperature in Kelvin

\Delta G= +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

Putting in the values:

\Delta G=(+ve)-T(-ve)

\Delta G=(+ve)(+ve)=+ve  

Reaction is non spontaneous at all temperatures.

4 0
3 years ago
The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14
Elena-2011 [213]

Answer:

Age of the atifact is 4.2\times 10^{2} years

Explanation:

  • For first -order radioactive decay- A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}
  • A_{t} represents activity of radioactive nuclide after t time, A_{0} represents initial activity of radioactive nuclide and t_{\frac{1}{2}} represents half-life
  • Here, A_{t}=19Bq, A_{0}=20Bq and t_{\frac{1}{2}}=5.73\times 10^{3}years

Plug-in all the given values in the above equation-

19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}

or, t=4.2\times 10^{2}

So, age of the atifact is 4.2\times 10^{2} years

6 0
3 years ago
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