A student measures the mass of an object and obtains three values: 21.57 g,

1. How many grams would 8.1 x 1021 molecules of sucrose (C12H22011) weigh?

Nimfa-mama [501]

Answer:

Mass = 4.6 g

Explanation:

Given data:

Number of molecules of sucrose = 8.1 ×10²¹ molecules

Mass of sucrose = ?

Solution:

First of all we will calculate the number of moles by using Avogadro number.

1 mole × 8.1 ×10²¹ molecules / 6.022×10²³ molecules

1.35 × 10⁻² mol

Mass of sucrose:

Mass = number of moles × molar mass

Molar mass = 342.3 g/mol

Mass = 1.35 × 10⁻² mol ×342.3 g/mol

Mass = 462.1 × 10⁻² g

Mass = 4.6 g

7 0

3 years ago

Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.

Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

$ln(K)=\frac{\Delta _RG}{RT}$

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

$Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}$

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

Best regards.

8 0

3 years ago

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HELP 20 points!

monitta

Answer:

The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.

Explanation:

The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.

5 0

2 years ago

WILL MARK BRAINLIEST!!!!

Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Compounds
Moles

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

Step 2: Identify Conversions

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})$
[DA] Divide [Cancel out units]: $\displaystyle 5.7015 \ mol \ K_2SO_4$

Step 4: Check

Follow sig fig rules and round. We are given 4 sig figs.

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

7 0

2 years ago

The weight of a table as measured by a student is 20 kg.The

miskamm [114]

(20/23) times 100 is 87%

now the answer asks for percentage ERROR

100-87 =13

so answer is C

7 0

2 years ago

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