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kaheart [24]
3 years ago
15

As part of your job you are asked to make 1 liter of a 0.5 molar sucrose solution. how much sucrose (c12h22o11) do you need? use

the information below listing the number of protons, neutrons, and electrons in each element.
Chemistry
1 answer:
kakasveta [241]3 years ago
6 0

Answer:-  171 g

Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.

We know that molarity is moles of solute per liter of solution.

If molarity and volume is given then, moles of solute is molarity times volume in liters.

moles of solute = molarity* liters of solution

moles of solute = 0.5*1 = 0.5 moles

To convert the moles to grams we multiply the moles by molar mass.

Molar mass of sucrose = 12(12) + 22(1) + 11(16)  

= 144 + 22 + 176

= 342 grams per mol

grams of sucrose required = moles * molar mass

grams of sucrose required = 0.5*342  = 171 g

So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.




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60.55%

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2 years ago
What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed
Ahat [919]

Answer:

             1.70 g.cm⁻³

Solution:

Data Given;

                   Mass  =  84.7 g

                   Volume  =  49.6 cm³

                   Density  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Putting values,

                   Density  =  84.7 g ÷ 49.6 cm³ 

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7 0
3 years ago
7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are
Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] =  0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

6 0
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