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2 years ago

According to the periodic table how many elements are metal

1 answer:

8 0

Around 95 of the 118 elements in the periodic table are metals. Examples of metal elements include iron, copper, silver, mercury, lead, aluminum, gold, platinum, zinc, nickel and tin.

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What are the relationships between melting and freezing of a mixture

Kaylis [27]

The temperature is the same but the heat flow is the opposite.

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2 years ago

Chemicals that are able to slow or stop the reaction rate are inhibitors. reactants. products. catalysts.

erma4kov [3.2K]

Answer:inhibitors

Explanation:

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2 years ago

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Coefficients are used in a chemical formula to show the number of each element. True or False

ANTONII [103]

Answer:

The same number of each element present before the reaction takes place must also be present on the product side of the equation. Coefficients are placed in front of a chemical formula to show the number of moles of that substances that are necessary for the reaction to occur.

Explanation:

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2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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3 years ago

Can anyone solve these balanced chemical equations

adell [148]

Answer: I have sent the notes to u private

Explanation:

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3 years ago

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