All categories

2 years ago

What are the transition metals

Chemistry

2 answers:

Valentin [98]2 years ago

8 0

Theyre the big bunched up group in the middle of the periodic table

Karolina [17]2 years ago

3 0

Transition metals are a large group of elements that include some of the most common metals such as copper and chromium. They also include elements that can create a magnetic field, such as nickel and iron. Transition metals are great conductors of electricity because the outer electrons are capable of flowing easily.

You might be interested in

You have 0.5 L of air at 203 k in an expandable container at constant pressure. You heat the container to 273 k. What is the vol

Jobisdone [24]

You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2

V1=0.5 L
T1=203 K
T2=273 K
V2=unknown

0.5L/203 = V2/273
V2= 0.67 L so C

Hope this helps :)

4 0

3 years ago

PLEASE HELP ME I NEED HELP

Crazy boy [7]

Answer:

5 moles S x (36.02 g S/mole S) = 180.1 grams of S

Explanation:

The periodic table has mass units for every element that can be correlated with the number of atoms of that element. The relationship is known as Avogadro's Number. This number, 6.02x $10^{23}$ , is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than 6.02x $10^{23}$ . The mole is correlated to the atomic mass of that element. The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units. But it can also be read as 36.02 grams/mole.

This means that 36.02 grams of S contains 1 mole (6.02x $10^{23}$ ) of S atoms.

This relationship holds for all the elements. Zinc, Zn, has an atomic mass of 65.38 AMU, so it has a "molar mass" of 65.38 grams/mole. ^5.38 grams of Zn contains 1 mole of Zn atoms.

And so on.

5.0 moles of Sulfur would therefore contain:

(5.0 moles S)*(36.02 grams/mole S) = 180.1 grams of S

Note how the units cancel to leaves just grams. The units are extremely helpful in mole calculations to insure the correct mathematical operation is done. To find the number of moles in 70 g of S, for example, we would write:

(70g S)/(36.02 grams S/mole S) = 1.94 moles of S. [Note how the units cancel to leave just moles]

4 0

2 years ago

Which of the following correctly describes the law of conservation of energy

egoroff_w [7]

Mass and energy can not be created or destroyed, they may be able to just be converted, and neither one seems without the opposite. For this reason in closed systems, both mass and energy are conserved individually. " I hope this helps "

3 0

3 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

No of moles = mass of the substance÷molar mass of the substance

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

Thus 52.8 g of CO2 is produced.

5 0

2 years ago

Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water

spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0

3 years ago