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Is Cr and Cu has same atomic radius?

Only the atomic radius is equivalent - 128 pm.


Is Chromium Cr Co C or Ch?

Chromium is denoted by the symbol Cr. It is a chemical element with atomic number 24.


How many grams of Co are there in a sample of Co that contains the same number of moles as a 45.6 gram sample of Cr?

The atomic weight of Cr is 52.0 g/mol1. Convert grams of Cr to moles of Cr:moles Cr = 45.6 g Cr1 mol = 0.877mol Cr52.0 gMultiply by moles per gram. Grams cancel out.The atomic weight of Co is 58.9 g/mol2. To convert 0.877 moles of Co to grams of Co:grams Co = 0.877 mol Co58.9 g = 51.7 g Co1 molMultiply by grams per mole. Moles cancel out.Note that since the atomic weight of Co is larger than the atomic weight of Cr, the mass of the same number of moles is also larger.


Where is Cr on the periodic table?

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What is the atomic mass of Cr?

Chromium has an atomic mass of 52.00 a.m.u.


What is the of protons in chromium?

There are 24 protons in Chromium (Cr). The atomic number of an element is the same as the number of protons.


How much protons does chromium have?

There are 24 protons in Chromium (Cr). The atomic number of an element is the same as the number of protons.


What is the atomis number of Cr?

Chromium is a metal. Atomic number of it is 24.


What elements have an ionic radius equal to the atomic radius?

a) Sc,Ti,V,Crb) Na,K,Rb,Csc) B,Si,As,Ted) F,Cl,Br,Ie) Na,Mg,Al,SiThe correct answer of these options is a) Sc,Ti,V,Cr because they are the closest elements to each other in the periodic table.


How many protons electrons and neutrons in chromium?

There are 24 protons in Chromium (Cr). The atomic number of an element is the same as the number of protons.


What does CR Co mean stamped on jewelry?

CR Co stamped on jewelry stands for Crown Jewelry Company, indicating the manufacturer or distributor of the piece.


What is the metal symbol for chromium?

Chromium has the chemical symbol Cr and an atomic number of 24 . Particulars can be found here: http://en.wikipedia.org/wiki/ChromiumWell, I have the same question respond please