Which state of matter undergoes changes in volume most easily

An engineer has created a new design for the pipe systems in homes. In this design, pipes are joined together with flexible conn

geniusboy [140]

Answer:

earthquakes

Explanation:

3 0

3 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

3 0

3 years ago

What is the empirical formula of a compound with a percent composition of 22.5% Phosphorous and 77.5% Chlorine?

sashaice [31]

Answer:

$\boxed {\boxed {\sf PCl_3}}$

Explanation:

We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.

We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.

22.5 g P
77.5 g Cl

Next, convert these masses to moles, using the molar masses found on the Periodic Table.

P: 30.974 g/mol
Cl: 35.45 g/mol

Use the molar masses as ratios and multiply by the number of grams. $22.5 \ g \ P * \frac {1 \ mol \ P }{30.974 \ g \ P}= \frac {22.5 \ mol \ P }{ 30.974} = 0.7264157035 \ mol \ P$

$77.5 \ g \ Cl * \frac {1 \ mol \ Cl }{35.45 \ g \ Cl}= \frac {77.5 \ mol \ Cl }{ 35.45} \ =2.186177715 \ mol \ Cl$

Divide both of the moles by the smallest number of moles to find the mole ratio.

$\frac {0.7264157035} {0.7264157035} = 1$

$\frac {2.186177715}{0.7264157035}=3.009540824 \approx 3$

The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as: PCl₃

4 0

2 years ago

For many weak acid or weak base calculations, you can use a simplifying assumption to avoid solving quadratic equations. Classif

shtirl [24]

Answer:

a. not valid

b. valid

c. not valid

d. valid

e. not valid

Explanation:

The assumption to avoid solving the quadratic equation for the calculation of [H⁺] and [OH⁻] involved in the equilibria of weak acids and bases ( small Ka and Kb) is valid as long as the value obtained from the shortcut is less than 5 % or less of the original acid or base concentration.

For a general monoprotic acid, as in this question, the equlibria is:

HA + H₂O ⇄ H₃O⁺ + A⁻ Ka = [H₃O⁺][A⁻]/[HA]

To determine the concentrations at equilibrium we are going to setupup the ICE table:

[HA] [H₃O] [A⁻]

Initial [HA]₀ 0 0

Change - x +x +x

Equil [HA]₀ - x x x

Ka = x² / [HA]₀ - x

Here is where we make our simplification of approximating [HA]₀ - x to the original acid concentration, [HA]₀, assuming x is much less than [HA] since HA is a weak acid.

To answer our questions we will solve for x,and then can compare it to the initial HA concentration.

Lets now perform our calculations.

(a) x = √ (0.01 x 1x 10⁻⁴) = 1 x 10⁻³ M = [H₃O⁺]

% = 1 x 10⁻³/.01 x 100 = 10%

The assumption is not valid.

(b) x = √ (0.01 x 1x 10⁻⁵) = 3.2 x 10⁻⁴ M = [H₃O⁺]

% = 3.2 x 10⁻⁴ /0.01 x 100 = 3.2 %

The assumption is valid since the criteria of 5 % or less has been met.

(c) x = √ (0.1 x 1x 10⁻³) = 1.0 x 10⁻² M = [H₃O⁺]

% = 1.0 x 10⁻² /0.1 x 100 = 10 %

The assumption is not valid, we wiould have to solve the quadratic equation.

(d) x = √ (1 x 1x 10⁻³) = 3.2 x 10⁻² M = [H₃O⁺]

% = 3.2 x 10⁻² / 1 x 100 = 3.2

The assumption is valid.

(e) x = √ (0.001 x 1x 10⁻⁵) =1.0 x 10⁻⁴ M = [H₃O⁺]

% = 1.0 x 10⁻⁴ / .001 = 10 %

The assumption is not valid and one has to solve the quadratic equation.

7 0

3 years ago

How many atoms of carbon are on the reactant side?

JulijaS [17]

Answer:I dont know

Explanation:

5 0

3 years ago