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3 years ago

Cuál es la masa molar del carbono?

Chemistry

1 answer:

maks197457 [2]3 years ago

6 0

Answer:

12.01 g

Explanation:

Si vas al simbolo de Carbono en la tabla periodica, cual es "C" ,el numero debajo del simbolo te dice que su masa molar es 12.01 g

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A highly polar molecule that contains a weak bond between a hydrogen atom and another element would be

sleet_krkn [62]

A highly polar molecule that contains a weak bond between a hydrogen atom and another element would be a strong acid.

8 0

4 years ago

When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is

ExtremeBDS [4]

Answer:

38.83 % of CaCO3

61.17 % of MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that...

CaCO3 molar mass = 100.09 g / mol = 100.09 x

MgCO3 molar mass = 84.31 g / mol = 84.31 y

decomposition reactions :

CaCO3 ---> CaO + CO2

MgCO3 ---> MgO + CO2

So we have that , Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3

thus

56.08 x + 40.30 y = 0.51 ( 100.09 x + 84.31 y )

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x

CaCO3 % in the sample

= 100.09 x × 100 / ( 100.09 x + 84.31 y )

= 10009 x / ( 100.09 x + 84.31 × 1.87 x )

= 10009 x / ( x ( 100.09 + 157.66 ) )

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83

= 61.17 %

7 0

3 years ago

2 Use the chart to answer the question. Solubility Curves of Various Salts 180 160 140 120 100 80 KNO3 in g NaCl in g KCl in g A

VMariaS [17]

According to the information in the graph, it can be inferred that the amount of solute that will precipitate out of solution at 20°C is 130 grams.

<h3>How to calculate the amount of solute that precipitates out of solution?</h3>

To calculate the amount of solute that precipitates out of solution we must identify the solute data at 80°C and 20°C and identify the difference as shown below:

Quantity of solute at 80°C: 170 grams.
Quantity of solute at 20°C: 40 grams.

170 grams - 40 grams = 130 grams

According to the above, the amount of solute that will precipitate out of solution due to the change in temperature is 130 grams of KNO3.

Note: This question is incomplete because the graph is missing. Here is the graph

Learn more about solute in: brainly.com/question/7932885

#SPJ1

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2 years ago

Which statement best describes nuclear binding energy

tia_tia [17]

When the protons and neutrons combine to form a nucleus, the mass that disappears

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4 years ago

Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen. determi

IceJOKER [234]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 70.79% = 70,79 g
H: 8.91% = 8.91 g
N: 4.59% = 4.59 g
O: 15.72% = 15.72 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol$

$H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol$

$N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol$

$O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}$

$H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}$

$N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}$

$O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}$

T<span>hus, the minimum or empirical formula found for the compound will be:

</span> $\boxed{\boxed{C_{18}H_{27}N_1O_3\:\:or\:\:C_{18}H_{27}NO_3}}\end{array}}\qquad\checkmark$

I hope this helps. =)

5 0

3 years ago