Which compound has the greatest percent composition by mass of sulfur

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

3 years ago

Which analogy can best be likened to the activation energy of a chemical reaction?

olganol [36]

The activation energy of a reaction is the minimum energy that must be overcome in order for the reaction to take place. One way of reaching the activation energy is by manipulating the process conditions like pressure or temperature. But the most common method is by adding an enzyme. An enzyme speeds up the rate of the reaction but does not actively take part in it.

An analogy would be pushing heavy wooden block down a slope. No matter how many people push on it, the block won't move because of friction. But if you spill oil on the floor, the block would effortlessly move down the slope. The oil here is like an enzyme in a reaction.

6 0

3 years ago

PLS HELP!!

Gnesinka [82]

Answer:

from the 1st equation:

4NH3 4NO

4 *(68) 4*30

1216 X mass of NO = 536.5 g

from the 2nd Equation

2NO 2NO2

2*30 2* 46

536.5 x mass of NO2 = 822.6 grams

from the 3rd Equation

3NO2 2HNO3

3*(46) 2* (63)

822.6 X mass of nitric acid = 751.06 gram

b) % yields = ( 96.2%* 91.3% *91.4%)= 80.3%

3 0

3 years ago

An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having

dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass 2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007 (1-x) = 1.008

2.014x + 1.007 - 1.007 x = 1.008

2.014x - 1.007x = 1.008 - 1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0

3 years ago

g Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane i

Kamila [148]

Answer:

6.05g

Explanation:

The reaction is given as;

Ethane + oxygen --> Carbon dioxide + water

2C2H6 + 7O2 --> 4CO2 + 6H2O

From the reaction above;

2 mol of ethane reacts with 7 mol of oxygen.

To proceed, we have to obtain the limiting reagent,

2,71g of ethane;

Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol

3.8g of oxygen;

Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol

If 0.0903 moles of ethane was used, it would require;

2 = 7

0.0903 = x

x = 0.31605 mol of oxygen needed

This means that oxygen is our limiting reagent.

From the reaction,

7 mol of oxygen yields 4 mol of carbon dioxide

0.2375 yields x?

7 = 4

0.2375 = x

x = 0.1357

Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g

8 0

3 years ago