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2 years ago

Suppose you analyze the composition of an unnamed compound. Your analysis shows that

2 answers:

6 0

First from the data given we are able to determine the empirical formula of the product. To do that we use the following algorithm:

1) Divide each percentage at the atomic mass of each element:

O 51.3 / 16 = 3.21

C 42.2 / 12 = 3.52

H 6.5 / 1 = 6.5

2) Now we divide the result to the lowest number, which is 3.21:

O 3.21 /3.21 = 1

C 3.52 /3.21 = 1.1 ≈ 1

H 6.5/3.21 = 2

So the empirical formula of the unknown compounds is:

CH₂O

We can conclude that the compound contains carbon, hydrogen and oxygen in the ratio given by the empirical formula:

C : H : O = 1 : 2 : 1

ale4655 [162]2 years ago

6 0

Answer : The compound is formaldehyde $CH_2O$ .

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.20 g

Mass of H = 6.50 g

Mass of O = 51.30 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.20g}{12g/mole}=3.52moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.50g}{1g/mole}=6.50moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{51.30g}{16g/mole}=3.21moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.52}{3.21}=1.09\approx 1$

For H = $\frac{6.50}{3.21}=2.02\approx 2$

For O = $\frac{3.21}{3.21}=1$

The ratio of C : H : O = 1 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_1H_2O_1=CH_2O$

Hence, from this we conclude that the compound is formaldehyde.

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A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

draw the organic product formed when 1−hexyne is treated with h2o, h2so4, and hgso4. click the draw structure button to launch t

V125BC [204]

The organic product formed when 1−hexyne is treated with H₂O, H₂SO₄, and HgSO₄ will be 2-hexanone (structure attached).

This reaction is an example of an oxymercuration reaction of the organic product 1−hexyne.

Oxymercuration is shown in three steps to the right. The nucleophilic double bond attacks the mercury ion, releasing an acetoxy group. The mercury ion's electron pair attacks carbon on the double bond, generating a positive-charged mercuronium ion. Mercury's dxz and 6s orbitals give electrons to the double bond's lowest unoccupied molecular orbitals.

In the second stage, the nucleophilic H₂O attacks the highly modified carbon, freeing its mercury-bonding electrons. Electrons neutralize mercury ions by collapsing. Water molecules have positive-charged oxygen.

In the third stage, the negatively charged acetoxy ion released in the first step attacks the hydrogen of the water group, generating the waste product HOAc. The two electrons in the oxygen-hydrogen link collapse into oxygen, neutralizing its charge and forming alcohol.

You can also learn about organic products from the following question:

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4 0

11 months ago

At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c

ivolga24 [154]

<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

<u>Initial:</u> 0.02 0.03

<u>At eqllm:</u> 0.02-2x x 0.03+x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}$

We are given:

$K_c=0.141$

Putting values in above equation, we get:

$0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135$

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

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3 years ago

How many grams in 82 moles of CO2

Tems11 [23]

44.0 g/mol

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3 years ago

Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie

aleksklad [387]

Explanation:

1. Water decomposition

Decomposition reactions are represented by-

The general equation: AB → A + B.

Various methods used in the decomposition of water are -

Electrolysis
Photoelectrochemical water splitting
Thermal decomposition of water
Photocatalytic water splitting

Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.

The chemical equation will be -

$H_2O + H_2 + O_2$

Hence, balancing the equation we need to add a coefficient of 2 in front of $H_2O$ on right-hand-side of the equation and 2 in front of $H_2$ on left-hand-side of the equation.