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mixer
16 hours ago
8

The label on a container of margarine lists "hydrogenated vegetable oil" as the major ingredient. Hydrogenated vegetable oil ___

_____.
Chemistry
1 answer:
KiRa [971]16 hours ago
5 0

Answer:

Hydrogenated vegetable oil maintains its solid state at room temperature.

Explanation:

Based on various chemists' studies I reviewed, it appears that hydrogenated vegetable oil contributes to margarine remaining solid at room temperature. Without it, the margarine would liquefy at room temperature, making it unsuitable or unsafe for consumption.

If you have further questions, feel free to reach out.

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A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal
Tems11 [846]

Answer:

A balloon will hold a gas volume of 45.0 L at a pressure of 1.35 atm and a temperature of 253 K.

Explanation:

Applying the Ideal Gas Law for the same gas amount, we have:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

Parameters are given as follows:

V₁ = 25.0 L

V₂ =?

P₁ = 2575 mm Hg

To convert pressure to atmospheres: P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using the earlier formula, we derive:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}

\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}

By solving for V₂, we find:

V₂ = 45.0 L

A balloon will hold a gas volume of 45.0 L at a pressure of 1.35 atm and a temperature of 253 K.

4 0
7 days ago
Classify these compounds as acid, base, salt, or other.? 1. ch3oh 3. hno3 5. nabr
KiRa [971]
<span> </span><span>1. Other (Alcohol)
3. Acidic
5. Salt

</span>
5 0
2 days ago
Read 2 more answers
A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74
KiRa [971]

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product (CH3COOH)^{2} gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol (CH3COOH)^{2} / 2 mol CH3COOH)

= 0.005 mol (CH3COOH)^{2}

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^{2}

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm

4 0
8 days ago
Which change of state is shown in the model?
Alekssandra [968]

I think the state change illustrated in the diagram is deposition. 
Deposition is the transformation of gases into solids without transitioning through a liquid phase. It is the reverse process of sublimation.
A key distinction between gases and solids lies in the spacing of molecules; gases have large spaces between molecules, whereas solids have very minimal spacing, resulting in solids being more densely packed. This is illustrated in the diagram showing the transition from gases to solids.

 
5 0
9 days ago
Read 2 more answers
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [956]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
14 days ago
Read 2 more answers
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