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3 years ago

15

A laboratory analysis of an unknown sample yields 40.5% carbon, 4.5% hydrogen, and 55.0% oxygen. if the sample has a molar mass

of 176.0 g/mol, what is the molecular formula of the sample

1 answer:

den301095 [7]3 years ago

6 0

Answer is: molecular formula is C₆H₈O₆.
n(C) = m(C) ÷ M(C).
n(C) = 40,5 g ÷ 12 g/mol.
n(C) = 3,375 mol.
n(H) = m(H) ÷ M(H).
n(H) = 4,5 g ÷ 1 g/mol.
n(H) = 4,5 mol.
n(O) = m(O) ÷ M(O).
n(O) = 55 g ÷ 16 g/mol.
n(O) = 3,4 mol.
n(C) : n(H) : n(O) = 3,375 mol : 4,5 mol : 3,4 mol / :3,375.
n(C) : n(H) : n(O) = 1 : 1,33 : 1.
n(C) : n(H) : n(O) = 3 : 4 : 3.
M(C₃H₄O₃) = 88 g/mol · 2 = 176 g/mol.

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When the volume of a gas is

Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

$\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

$T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

The units of milliliters (mL) cancel.

$T_2=\frac{ 137 \textdegree C * 425 }{250 }$

$T_2= \frac{58225}{250} \textdegree C$

$T_2=232.9 \textdegree C$

The temperature changes to <u>232.9 degrees Celsius.</u>

3 0

3 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

How many moles of oxygen are necessary to generate 28 moles of water, according to the following equation: 2H2+O2→2H2O

valentinak56 [21]

The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles

<h3>Balanced equation </h3>

2H₂ + O₂ —> 2H₂O

From the balanced equation above,

2 moles of water were obtained from 1 mole of oxygen

<h3>How to determine the mole of oxygen needed </h3>

From the balanced equation above,

2 moles of water were obtained from 1 mole of oxygen

Therefore,

28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen

Thus, 14 moles of oxygen are needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

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2 years ago

How many grams of nitrogen dioxide are required to produce 5.89x10^3 kg of hno3 in excess water

polet [3.4K]

<span>3 NO2 + H2O -------->. 2 HNO3. + NO
3(46g)------------------------> 2 ( 63g) HNO3
? kg-------------------------5.89 x10^3kg HNO3
Mass of NO2. = 5.89x10^3 x 138/ 2(63) = 6.45 x10^3 kg</span>

5 0

3 years ago

Choose the best tool for each scenario.

Mama L [17]

Answer:

a) Graph

b) Weight balance or gas syringe or upside-down measuring cylinder

Explanation:

a) Identifying a trend in temperature change over time - The best tool for this scenario is to represents the temperature daily, weekly, monthly or annually on graph to interpret the fluctuation in temperature owing to local seasonal changes and weather conditions

b) Measuring the mass of a product of a chemical reaction - If the product is solid or liquid then the balance is used to measure the mass. If the product is a gas, then gas syringe or upside-down measuring cylinder is used.

8 0

3 years ago