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Kruka [31]
3 years ago
7

What is the oxidation number of chromium in dichromate ion, cr2o72-?

Chemistry
1 answer:
castortr0y [4]3 years ago
8 0
Oxidation number is charge of element in compound. Can be neutral, positive or negative.
Oxygen in dichromate has oxidation number -2, becauce there are seven oxygens, net oxidation number of oxygen is -14.
2·x(oxidation number of Cr) + 7· (-2) = -2   2x= +12
x= +6, oxidation number of one chromium is +6.
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How many kJ of energy will be released when 4.72g of carbon react with excess oxygen to produce carbon dioxide (delta H is -393.
ivolga24 [154]

Answer:

155 kJ of energy will be released.

Explanation:

The \rm \Delta H\textdegree (\rm \Delta H \textdegree_\text{rxn} in some textbooks) here stands for standard enthalpy change per mole reaction. To find the amount of energy released in this reaction, start by finding the number of moles of this reaction that will take place.

How many <em>moles</em> of atoms in 4.72 grams of carbon?

Relative atomic mass data from a modern periodic table:

  • C: 12.01.

\displaystyle n = \frac{m}{M} = \rm \frac{4.72\;g}{12.01\; g\cdot mol^{-1}} = 0.393006\; mol.

The coefficient of carbon in the equation is one. In other words, each mole of the reaction will consume one mole of carbon. Oxygen is in excess. As a result, \rm 0.393006\; mol of carbon will support \rm 0.393006\; mol of the reaction.

How much energy will be released?

The \rm \Delta H\textdegree{} value here is negative. But don't panic. \rm \Delta H\textdegree{} is the same as the chemical potential energy of the reactants minus the products in one mole of the reaction. \rm \Delta H\textdegree{} = -393.5\;kJ means that the chemical potential energy drops by \rm 393.5\; kJ during each mole of the reaction (with the coefficients as-is.) Those energy difference will be released as heat. In other words, one mole of the reaction will release \rm 393.5\;kJ of energy.

The 4.72 grams of carbon will support \rm 0.393006\; mol of this reaction. How much heat will that \rm 0.393006\; mol of reaction release?

Q = n \cdot (-\Delta \text{H}\textdegree{}) = \rm 0.393006\times 393.5 = 155\;kJ.

As a side note, the mass of carbon 4.72 grams is the least significant data in this question. There are three significant figures in this value. As a result, keep more than three significant figures in calculations but round the final result to three significant figures.

7 0
4 years ago
Which process produces the energy that is used in photosynthesis?
Svetlanka [38]

Answer:

chemical reactions

Explanation:

the sunlight is converted to energy

8 0
3 years ago
A<br>Which property of a gas affects the rate at which it spreads throughout<br>laboratory?​
balu736 [363]

Answer:

Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. Effusion refers to the movement of gas particles through a small hole. Graham's Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles

5 0
3 years ago
217 mL of neon gas at 551 celsius is brought to standard temperature (0 celsius) while the pressure is held constant. What is th
pantera1 [17]

Answer: The new volume is 72 ml

Explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=217ml\\T_1=551^oC=(551+273)K=824K\\V_2=?\\T_2=0^0C=(0+273)K=273K

Putting values in above equation, we get:

\frac{217ml}{824K}=\frac{V_2}{273K}\\\\V_2=72ml

Thus the new volume is 72 ml

7 0
2 years ago
When parallel light rays exit a concave lens,the light ray is?
ad-work [718]
When parallel rays exit a concave lens, the light rays are divergent.
The rays diverge or bend away from the axis it has been traveling upon entering the lens when it reaches the other side of the lens. These rays appear to have come from the same focal point before entering the concave lens. When these parallel rays are extended, it will be traced back to a single point of origin.
5 0
3 years ago
Read 2 more answers
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