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1 year ago

Molecules containing a large number of hydroxyl groups are

1 answer:

6 0

Molecules containing a large number of hydroxyl groups are soluble in water. Molecules which contains the functional group hydroxyl or -OH are called alcohols. Water and alcohols share some similar properties. One would be its polar properties. Alcohols are polar molecules as they contain a partial positive and partial negative end. Making them to be miscible in water. The hydroxyl groups in alcohol are capable of forming hydrogen bonds with the water molecules and with other alcohols. As the number of the hydroxyl groups increases, the polarity of the molecule would also increase making it more soluble in water.

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a mixture was found to contain 1.05g of sio2, 0.69g of cellulose, and 1.82g of calcium carbonate, what percentage of calcium car

enot [183]

Mass percentage is another way of expressing concentration of a substance in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%. It is calculated as follows:

% CaCO3 = (<span>1.82g of calcium carbonate</span> / (1.05 g SiO2 + 0.69 g of cellulose + <span>1.82g of calcium carbonate)) x 100% = 51.12% Calcium carbonate</span>

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1 year ago

Which of the following actions cannot induce voltage in a wire?

MariettaO [177]

Your answer is D. Since there is little to no magnetic field to wire, if it is copper which most wires are, there will be no voltage in a wire.

7 0

2 years ago

Read 2 more answers

The combustion of propane is represented by the following chemical equation. C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l) The standard enthalp

wariber [46]

Answer:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

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2 years ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0

2 years ago

An ice cube measuring 5.80 cm by 5.80 cm by 5.80 cm has a density of 0.917 g/mL What is the mass?

OLga [1]

Answer: 178.9 g

Explanation:

Density = $\frac{mass}{volume in mL}$

find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³

1.0 cm³ = 1.0 mL

so 195.112 cm³ = 195.112 mL

plug value into density equation:

0.917 g/mL = (mass) / (195.112 mL)

and solve for mass!

3 0

1 year ago