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zloy xaker [14]
3 years ago
5

Help me please ...online claases got me crazy

Chemistry
1 answer:
White raven [17]3 years ago
6 0

Answer:

1,000 -  2,000

Explanation:

Just look at the bar graph. The bar ends at about 1,500 which is between 1,000 - 2,000. That is the only valid answer for this problem.

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Hydrogen reacts with chlorine to form hydrogen chloride (HCl (g), mc012-1.jpgHf = –92.3 kJ/mol) according to the reaction below.
Kobotan [32]

<u>Given:</u>

H2(g) + Cl2 (g) → 2HCl (g)

<u>To determine:</u>

The enthalpy of the reaction and whether it is endo or exothermic

<u>Explanation:</u>

Enthalpy of a reaction is given by the difference between the enthalpy of formation of reactants and products

ΔH = ∑nHf (products) - ∑nHf (reactants)

      = [2Hf(HCl)] - [Hf(H2) + Hf(Cl2)] = 2 (-92.3) kJ = - 184.6 kJ

Since the reaction enthalpy is negative, the reaction is exothermic

<u>Ans:</u> The enthalpy of reaction is -184. kJ and the reaction is exothermic

3 0
3 years ago
Read 2 more answers
Calculate the radius of a tantalum (ta) atom, given that ta has a bcc crystal structure, a d ensity of 16.6 g/cm3, and an atomic
Sidana [21]

solution:

Ta is BCC, a=\frac{4}{\sqrt{3}r},n=2atoms/cell\\p=16.6gram/cm^3,A=180.9gram/mol\\P=\frac{A\times n}{N_{a}\times v}=\frac{180.9\frac{gram}{mol}\times 2 \frac{atom}{cell}}{6.023\times10^23\frac{atom}{mol}\times(\frac{4}{\sqrt{3}\times r})^3\frac{cm^3}{cell}}\\=16.6\frac{gram}{cm^3}\\r=(\frac{180.9\times2}{16.6\times6.023\times10^23\times(\frac{4}{\sqrt{3}})^3})^\frac{1}{3}\times10^8


4 0
3 years ago
• 1) The pressure of a sample of gas in a 2.00-L container is 876 mmHg.
kifflom [539]

Answer:

4380 mmHg

Explanation:

Boyle's Law can be used to explain the relationship between pressure and volume of an ideal gas. The pressure is inversely related to volume, so if volume decrease the pressure will increase. It can be expressed in the equation as:

P1V1=P2V2

In this question, the first condition is 2L volume and 876 mmHg pressure. Then the system changed into the second condition where the volume is 400ml and the pressure is unknown. The pressure will be:

P1V1= P2V2

876 mmHg * 2L = P2 * 400ml /(1000ml/L)

P2= 876 mmHg * 2L / 0.4L

P2= 4380 mmHg

5 0
3 years ago
at standard pressure how do the boiling point and the freezing point of NaCl compare to the boiling point and freezing point of
Ede4ka [16]
Usually in this context you would be referring to the boiling and freezing point of a NaCl <em>solution</em> (saltwater) compared to pure H_{2}O. Sematics would be different for NaCl compound itself, you would say melting and boiling point for a solid substance- and the temperatures would be very, very radical (high). 
The boiling point of pure water is 100 degrees C (212 F), and the freezing/melting point is below 0 degrees C (32 F). For a salt water solution, the boiling point is raised and the melting point is lowered. This means that water will stay liquid for an increased range of temperature. Depending on the amount of NaCl solute in the water, the boiling and melting points may change a few degrees.
7 0
3 years ago
The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

Now put all the given values in this expression, we get:

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

7 0
3 years ago
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