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3 years ago

Draw an ether that contains exactly five carbon atoms and only single bonds

Chemistry

1 answer:

Eddi Din [679]3 years ago

8 0

Answer:

H3CH2COCH2CH2CH3

Ether:

Ether are organic compounds that contain ether functional group , in which oxygen atom is connected with two alkyl or aryl group.

They have general formula as follow

R---O---R or R'---O----R or R'---O---R'

while R = Alkyl

R' = Aryl

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A compound containing only carbon, hydrogen, and oxygen is analyzed using combustion analysis. When 50.1 g of the compound is bu

juin [17]

Answer:

$C_{3}H_4O_2$

Explanation:

Hello,

In this case, since the carbon of the initial compound is present in the carbon dioxide product, we can compute the mass and moles of carbon in the compound:

$n_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =2.09molC\\\\m_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2}*\frac{12gC}{1molC} =25.0gC$

Next, the mass and moles of hydrogen in the compound, is contained in the yielded amount of water, thus, we compute the mass and moles of hydrogen in the compound:

$n_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =2.79molH\\\\m_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} *\frac{1gH}{1molH} =2.79gH$

In such a way, the mass of oxygen comes from the mass of the compound minus the mass of carbon and oxygen:

$m_O=50.1g-25.0g-2.79g=22.31gO$

And the moles:

$n_O=22.31gO*\frac{1molO}{16gO}=1.39molO$

Then, we compute the subscripts by diving the moles of C, H and O by the moles of oxygen as the smallest moles:

$C:\frac{2.09}{1.39}=1.5 \\\\H:\frac{2.79}{1.39}=2\\ \\O:\frac{1.39}{1.39} =1$

After that, we write:

$C_{1.5}H_2O$

Which must be shown in whole number only, thereby we multiply the subscripts by 2, so the empirical formula turns out:

$C_{3}H_4O_2$

Best regards.

5 0

3 years ago

The equilibrium constant, Kp, for the following reaction is 9.52Ã10-2 at 350 K: CH4(g) + CCl4(g) 2CH2Cl2(g). Calculate the equil

Nimfa-mama [501]

Answer:

A) p CH₄ = 0.732 atm

B) p CCl₄ = 0.732 atm

c) p CH₂Cl₂ = 0.22 atm

Explanation:

we have the equilibrium constant for this problem, along the initial pressures for the reactants, and need to find the partial pressures at equilibrium. So lets setup the equilibrium:

CH₄ (g) + CCl₄ (g) ⇔ 2 CH₂Cl₂ (g) Kp = 9.50 x 10⁻²

where Kp is given by :

Kp = p CH₂Cl₂ ² / p CH₄ x p CCl₄

where p are the partial pressures

p CH₄ atm p CCl₄ atm p CH₂Cl₂ atm

initial 0.844 0.844 0

change - x - x +2x

equilibrium 0.844 - x 0.844 - x 2 x

Kp = 9.52 x 10⁻² = ( 2x )²/ (( 0.844 - x ) x ( 0.844 - x ))

9.52 x 10⁻² = (2x)² / ( 0.844 - x )²

Taking square root to both sides of the equation:

√9.52 x 10⁻² = 2x / (0.844 - x )

0.309 = 2x / (0.844 - x)

0.260 - 0.309 x = 2x

0.260 = 2.309 x ⇒ x = 0.112

So the partial pressures are at equilibrium are:

p CH₄ = p CCl₄ = 0.844 - 0.112 = 0.732 atm

p CH₂Cl₂ = 2 x (0.112 atm) = 0.224 atm

You can check your answer is correct by plugging this values and comparing with the given Kp.

6 0

3 years ago

Read 2 more answers

Q10.An aqueous solution contains 5.00x10-2 mol/L of Ca2+ and 7.00x10-3 mol/L of SO4

romanna [79]

Answer:

The precipitate will form.

Explanation:

Let's write the equilibrium expression for the solubility product of calcium sulfate:

$CaSO_4(s)$ ⇄ $Ca^{2+}(aq)+SO_4^{2-}(aq)$

The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:

$K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}$

Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:

$Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}$

Now given the K value of:

$K_{sp}=10^{-4.5}=3.2\cdot10^{-5}$

Notice that:

$Q_{sp}>K_{sp}$

This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.

6 0

4 years ago

to save time you can approximate the initial volume of water to ±1 ml and the initial mass of the solid to ±1 g . for example, i

makkiz [27]

The correct answer for this question is

20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.

15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.

11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.

To know more about Density, click here:

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8 0

2 years ago

The chemical formula of aspirin is C9H8O4. What is the mass of 0.40 mol of aspirin?

dybincka [34]

Answer:

$\boxed {\boxed {\sf About \ 72 \ grams }}$

Explanation:

To convert from moles to grams, we must use the molar mass (also known as the gram formula mass).

First, look up the molar masses of the elements in the formula.

C: 12.011 g/mol
H: 1.008 g/mol
O: 15.999 g/mol

Next, multiply by the subscript, because it tells us the number of atoms of each element in the formula.

C₉: 4(12.011 g/mol)= 108.099 g/mol
H₈: 8(1.008 g/mol)= 8.064 g/mol
O₄: 4(15.999 g/mol)= 63.996 g/mol

Add the values.

108.099 + 8.064+63.996=180.159 g/mol

Use this molar mass as a ratio.

$\frac {180.159 \ g \ C_9H_8O_4}{1 \ mol \ C_9H_8O_4}$

Multiply by the given number of moles, 0.40

$0.40 \ mol \ C_9H_8O_4 *\frac {180.159 \ g \ C_9H_8O_4}{1 \ mol \ C_9H_8O_4}$

The units moles of aspirin cancel.

$0.40 *\frac {180.159 \ g \ C_9H_8O_4}{1 }$

$72.0636 \ g \ C_9H_8O_4$

The original number of moles has 2 sig figs (4 and 0), so answer must have the same. For the number we calculated, it is the ones place. The 0 in the tenth place tells us to leave the 2.

$72 \ g$

5 0

3 years ago