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3 years ago

This is how manganese appears in the periodic table.

Chemistry

1 answer:

Allisa [31]3 years ago

6 0

Answer:

Atomic Symbol

Explanation:

States which atom it is as an abbreviated version

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ethanol is a common laboratory solvent and has a density of 0.789 g/mL. what is the mass, in grams, of 151 mL of ethanol

natta225 [31]

V = 151 mL = 151 cm³
d = 0,789 g/mL = 0,789 g/cm³
--------------------------------------

d = m/V
m = d×V
m = 0,789×151
m = 119,139g

3 0

4 years ago

Read 2 more answers

In terms of bonds, what would the molecule C₆H₁₂ be classified as?

Colt1911 [192]

Answer:

Alkene

Explanation:

5 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago

The density of gold is 19.3 g /cm cubed The density of iron pyrite is 5.0 g /cm cubed. Is a nugget of iron pyrite and a nugget o

telo118 [61]

If you clear volume in the density equation:

$\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}$

The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.

$V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3$

$V_{iron} = \frac{m}{\rho} = \frac{50\ g}{5.0\ g/cm^3} = \bf 10\ cm^3$

6 0

4 years ago

A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water

Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be $(C_6H_{12}O)_n$

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0

3 years ago