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2 years ago

What mass of sucrose needs to be dissolved in 540 g of water in order to prepare a 15% by mass solution.

Chemistry

1 answer:

Lilit [14]2 years ago

5 0

Answer:

95.3g

Explanation:

Given parameters:

Mass of water = 540g

Percentage by mass of solution = 15%

Unknown:

Mass of sucrose = ?

Solution;

To solve this problem, we must understand that a solution is made up of solute dissolved in a solvent. A solute is the solid dissolved in another. The solvent is the is dissolving medium;

Percentage by mass of solute in solution = $\frac{Mass of solute}{Mass of solution} x 100$

Mass of solution = mass of water + Mass of sucrose = 540 + M

where M is the mass of sucrose;

therefore;

$\frac{M}{540 + M } x 100$ = 15

Solving for M gives 95.3g

Mass of sucrose is 95.3g

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He is most well known for discovering the role Oxygen plays in combustion

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3 years ago

Describe un modelo matemático para representar lo que sucede en cada una de las reacciones químicas (incluye toda la simbología

GuDViN [60]

Responder:

2H2 + O2 → 2H2O

CaO + H2O → Ca (OH) 2

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Explicación:

2H2 + O2 → 2H2O

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2 years ago

If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

Consider the following reaction: 

SO2Cl2 -----> SO2(g) + Cl2(g) 

Kc= 2.99 x 10^-7 at 227 degrees celcius 

If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

SOCl2 ---------> SO2 + Cl2 Kc = 2.99x10⁻⁷

i) 0.168 0 0

c) -x +x +x

e) 0.168-x x x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

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[Cl₂] = 2.24x10⁻⁴ M

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3 years ago

El sol es la estrella más grande

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Answer:

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3 years ago

The titration of a 20.0-mLmL sample of an H2SO4H2SO4 solution of unknown concentration requires 22.87 mLmL of a 0.158 M KOHM KOH

kkurt [141]

Answer:

0.0905 M

Explanation:

Let's consider the neutralization reaction between H2SO4 and KOH.

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

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0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol

1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:

M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M

3 0

3 years ago