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Rasek [7]
3 years ago
8

What is the molality of a solution prepared by dissolving 150.0 g C6H12O6 in 600.0 g of H2O?

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

50/36 = 25/18

Explanation:

Solution at attachment box

Molality = mole of dissolvable (this question glucose) / kg of water

Masja [62]3 years ago
3 0

Answer:

Molality = 1.38 mol/Kg

Explanation:

Given data:

Molality of solution = ?

Mass of C₆H₁₂O₆ = 150.0 g

Mass of water = 600.0 g (600 g ×1 kg/1000 g= 0.6 Kg)

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 150.0 g/180.16 g/mol

Number of moles = 0.83 mol

Molality:

Molality = moles of solute / kg of solvent

Molality =  0.83 mol /0.6 Kg

Molality = 1.38 mol/Kg

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2Na2SO4 :How many atoms for each element are in the formula?<br><br> S=<br> O=<br> Na=
yuradex [85]

Answer:

S=  2(1)  = 2

O=  2(4)  = 8

Na= 2(2)  = 4

Explanation:

The given compound is:

          2Na₂SO₄  

An element is a distinct substance that cannot be split up into simpler substances.

So;

  Number of atoms of elements here are:

S=  2(1)  = 2

O=  2(4)  = 8

Na= 2(2)  = 4

6 0
2 years ago
What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0
3 years ago
10 To become positively charged, an atom must: A Galn a proton B Lose a proton C С. Galn an electron D Lose an electron } ​
Studentka2010 [4]

Answer:

D: lose an electron

Explanation:

when an atom loses an electron it's positively charged and when it gain an electron it is negatively charged

6 0
3 years ago
Two copper electrodes dipped in copper sulphate solution are connected to the end of the battery marked with a ''_'' is:
guajiro [1.7K]

Answer:

d) catión

Explanation:

8 0
3 years ago
Read 2 more answers
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
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