Anyone else ever done this? Please help !!

ХА

kvasek [131]

Answer:

B. Biosphere.

Explanation:

8 0

3 years ago

Draw the structures of the aldol addition and condensation products of 3‑methylbutanal

Leona [35]

Aldol condensation involves two aldehydes or two ketones or an aldehyde and a ketone. The product of the reaction is shown in the image attached.

<h3>What is Aldol condensation?</h3>

Aldol condensation is a reaction that involves two aldehydes or two ketones or an aldehyde and a ketone. This reaction occurs at the carbonyl group as one of the carbonyl compounds is in the enolate form.

The product of the reaction is shown in the image attached. This gives the mechanism for the Aldol condensation of the compound 3‑methylbutanal.

Learn more about aldol condensation: brainly.com/question/9415260

6 0

2 years ago

The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M

anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0

3 years ago

Read 2 more answers

Of the choices below, which is true for the relationship shown? It is Ka for the acid H3P2O72−. It is Kb for the acid H3P2O72−.

Naily [24]

The expression for the Ka for the given acid is:

Ka = [H2P2O7^2-] [H3O+] /[H3P2O7^2-]

<span>Ka is the acid dissociation constant or the acidity constant. It is a measure of the acid strength when in solution. It is an equilibrium constant for the dissociation of the acid.</span>

3 0

3 years ago

Read 2 more answers

Determine the empirical formula of a compound containing 40. 6 grams of carbon, 5. 1 grams of hydrogen, and 54. 2 grams of oxyge

zavuch27 [327]

The empirical formula is C₂H₃O₂

<h3>What is Empirical formula of a compound ?</h3>

The empirical formula is the simplest whole number ratio of elements present in a compound.

The total molar mass of the compound is 118.084 g/mol.

mass of Carbon present = 40.6

mass of Hydrogen present = 5.1 grams

mass of Oxygen present = 2 grams

Moles of C = 40.6/12 = 3.38

Moles of H = 5.1/1.008 = 5

Moles of Oxygen = 54.2/15.999 = 3.38

Ratio of Moles of C to Oxygen is 1 : 1

Ratio of Moles of C to H is 1/1.5

Multiplying each mole fraction by 2

The empirical formula is C₂H₃O₂

To know more about Empirical Formula

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5 0

2 years ago