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Alja [10]
3 years ago
15

Sugar is easily soluble in water and has a molar mass of 342.30 g/mol. What is the molar concentration of a 259.8 mL aqueous sol

ution prepared with 73.1 g of sugar?
Chemistry
1 answer:
DanielleElmas [232]3 years ago
5 0

Answer:

M=0.822M

Explanation:

Hello,

In this case, given the mathematical definition of molar concentration as the moles of solute divided by the liters of solution:

M=\frac{n}{V}

We can easily compute it by firstly computing the moles of sugar by using its molar mass:

n=73.1g*\frac{1mol}{342.30g} =0.214mol

After that, we compute the concentration by realizing 259.8 mL equals 0.2598 L:

M=\frac{0.214mol}{0.2598L}\\ \\M=0.822mol/L

Or just:

M=0.822M

Considering molar units as mol/L

Best regards.

You might be interested in
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
Write the lewis structure for mgi2. draw the lewis dot structure for mgi2. include all lone pairs of electrons.
viktelen [127]
Lewis Structure is drawn in following steps,

1) Calculate Number of Valence Electrons:
    
# of Valence electrons in Mg  =  2
# of Valence electrons in I      =  7
# of Valence electrons in I      =  7
                                               ---------
Total Valence electrons          =  16

2) Draw Mg as a central atom surround it by two atoms of Iodine.

3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.

Total Valence electrons           16
- Four electrons                    -   4
                                              ----------
                                                  12

4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.

The final lewis structure formed is as follow, 

4 0
3 years ago
Convert 90.23 kPa into atm
svlad2 [7]

Answer:

The answer is

<h2>0.89 atm </h2>

Explanation:

To convert from kPa to atm we use the conversion

101.325 kPa = 1 atm

If

101.325 kPa = 1 atm

Then

90.23 kPa will be

\frac{90.23}{101.325}  \\  =0.89050086...

We have the final answer as

<h3>0.89 atm</h3>

Hope this helps you

6 0
3 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

6 0
3 years ago
Read 2 more answers
Name the compound mg3cl2 is it magnesium chloride?
Scilla [17]
No 
Magnesium Chloride is MgCl2 

The method I use to name ionic compounds is 'swap and drop' 

Mg oxidation number is +2 and Cl oxidation number is -1
Mg^2+         Cl^-1  
'swap'
Mg^1           Cl^2 
'drop' 
MgCl2
 
7 0
3 years ago
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