What is the most common ion charge of chromium? Where on the periodic table?

Chemistry

2 answers:

raketka [301]2 years ago

7 0

Answer:

Explanation:

valina [46]2 years ago

7 0

The most common oxidation states of chromium are +6, +3, and +2. On the metal area of the periodic table.

Explanation:

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Is elemental phosphorus dull or high luster?

Ugo [173]

Each element can usually be classified as a metal or a non-metal based on their ... They are usually dulland therefore show no metallic luster and they do not reflect ... Dull, Brittle solids; Little or no metallic luster; High ionization energies; High ...

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3 years ago

How to write a balanced chemical equation from empirical formula?

netineya [11]

Let's use the example: H2O ---> H2 + O2
We find how many elements of a product are on one side and how many elements on the other side.
Reactant: H=2 O=1
Product: H=2 O=2
We need to make the same amount of hydrogen and oxegyn atoms on each side, regardless of how high the numbers are, and we do this by adding coefficients to the compounds.

Reactant: H=4 O=2
Product : H=4 O=2
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3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$