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victus00
2 months ago
5

How many liters of cane juice is needed to supply 5g sucrose if cane juice contains 12% sucrose

Chemistry
1 answer:
alisha [2.9K]2 months ago
6 0

Given parameters:

Mass of sucrose  = 5g

Density of sucrose  = 1.12g/mL

Percentage of sucrose per liter of cane juice  = 12%

Unknown:

Volume of cane juice required =?

We need to understand the relationship between volume and density. Density represents mass per unit volume.

Mathematically;

                Density  = \frac{mass}{volume}

Now, calculate the volume of sucrose;

                  1.12g/mL = \frac{5}{Volume}

       Volume  = \frac{5}{1.12}    =  4.46mL   = 4.46 x 10⁻³L since 1000mL  = 1L

Since 12% of one liter of cane juice is sucrose,  

                    12% of  x  liter of cane juice  = 4.46 x 10⁻³L

                   Volume of cane juice  = 4.46 x 10⁻³ x \frac{100}{12}   = 0.037L

Volume of cane juice needed is  0.037L

   

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Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

  • The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
  • Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}

The molarity of Cu²⁺ is:

\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M

4 0
2 months ago
If Sara was planning a wedding and wanted to have a sculpture of a heart made of butter at the reception, describe how Sara coul
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inspect the packaging

Explanation:

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2 months ago
For which of the following reactions is ΔHrxn equal to ΔHf of the product? You do not need to look up any values to answer this
Alekssandra [3086]

Answer:

In all listed reactions, ΔH°rxn does not correspond to the ΔH°f of the resulting product.

Explanation:

The standard enthalpy of formation (ΔH°f) signifies the enthalpy change that occurs when 1 mole of a product is created from its basic elements in their standard states.

1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)

ΔH°rxn does not equal ΔH°f of the product, since H₂O(g) is a compound rather than an element.

Na⁺(g) + F⁻(g) ⟶ NaF(s)

ΔH°rxn is not the same as ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).

K(g) + 1/2 Cl₂(g) ⟶ KCl(s)

ΔH°rxn is not equal to ΔH°f of the product due to K being outside its standard state (K(s)).

O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)

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Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.
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vHe / vNe = 2.24

Explanation:

To determine the velocity of an ideal gas, one should apply the formula:

v = √3RT / √M

In this equation, R represents the gas constant (8.314 kgm²/s²molK); T refers to temperature, and M indicates the molar mass of the gas (4x10⁻³kg/mol for helium and 20.18x10⁻³ kg/mol for neon). Hence:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio simplifies to:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it assists you!

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