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Anestetic [448]
4 years ago
6

Since the new oceanic crust is created at the mid-ocean ridges, then the Earth is getting bigger.

Chemistry
1 answer:
Svet_ta [14]4 years ago
7 0

Answer:

False

Explanation:

It is true that new crust is being created at the mid-oceanic ridges, but this does not make the Earth any bigger despite the the growth of oceanic crusts in these regions.

The reason that Earth does not get bigger is because the same plates that are being created at the divergent plate boundaries are recycled into the mantle at the subduction zones. Because of this, the surface area of the Earth remains the same.

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A shiny chunk of metal is found to have a mass of 37.28g. The metal is dropped into a graduated cylinder which contains 20.0 mL
Amanda [17]

Answer: The density of the material is 2.66 g/mL and it is likely this  is made of Aluminum

Explanation:

The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = \frac{m (mass)}{ v (volume)}. Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:

P = \frac{37.28g}{14.0mL}

P = 2.66 g/mL

This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.

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4 years ago
Amount of valence electrons KMnO4
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3 years ago
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4 years ago
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What’s the electron configuration of an N-3 ion
Alla [95]

Answer:

1s2 2s2 2p6

Explanation:

The nitride ion(N^3-) is formed when nitrogen gains three electrons. Nitrogen possesses seven electrons in its orbitals and ordinarily has the electronic configuration; 1s2 2s2 2p3. However,being in group 15, nitrogen can accept three electrons to form the nitride ion and complete its octet of electrons. When this happens, three electrons are added to the nitrogen atom and the electronic configuration is now the same as that of Neon, its closest noble gas which is 1s2 2s2 2p6. Hence the answer given above.

Elements can accept or donate electrons in order to complete their octet structure in accordance to the octet rule which states that atoms and ions must possess eight electrons in their outermost shell in order to attain chemical stability. The reason for ion formation and chemical reaction is in order for species to attain the octet structure.

8 0
3 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
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