Can someone help me with this question?

Why was mass lost from the crucible during the reaction?

Alexxandr [17]

If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.

6 0

3 years ago

Us

MrRissso [65]

Answer:

24.9%

Explanation:

According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:

0.0927 × 100 = 9.27%

Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;

100% - 9.27% = 90.73%

THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles

N.B: mole = mass/molar mass

Given the Molar Mass

NaCl: 58.44 g/mol

H2O: 18.016 g/mol

For NaCl;

0.0927 = mass/58.44

mass = 0.0927 × 58.44

5.42g

For H2O;

9.073 = mass/18.016

mass = 9.073 × 18.016

= 16.35g

Total mass of solution = 16.35g + 5.42g = 21.77g

Mass percent of NaCl = mass of NaCl/total mass × 100

% mass of NaCl = 5.42g/21.77g × 100

= 0.249 × 100

= 24.9%

5 0

3 years ago

Read 2 more answers

How many molecules are in 2.50 moles of carbon dioxide?

podryga [215]

Answer:

1.5055×10²⁴ molecules

Explanation:

From the question given above, the following data were obtained:

Number of mole CO₂ = 2.5 moles

Number of molecules CO₂ =?

The number of molecules present in 2.5 moles CO₂ can be obtained as:

From Avogadro's hypothesis,

1 mole of CO₂ = 6.022×10²³ molecules

Therefore,

2.5 mole of CO₂ = 2.5 × 6.022×10²³

2.5 mole of CO₂ = 1.5055×10²⁴ molecules

Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂

7 0

3 years ago

All elements can be classified as metals, metalloids, or ?

svlad2 [7]

All elements can be classified as metals, metalloids, or nonmetals

5 0

3 years ago

For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)

juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene

Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

Mass CO2 = 9.15 grams

4 0

3 years ago