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3 years ago

Lipids with high polyunsaturatd fatty acid content are ____________ at room temperature.

1 answer:

7 0

The answer to this question would be liquid.

Lipids with high polyunsaturated fatty acid mostly found as a liquid, but the saturated one would be solid at room temperature. A molecule with higher molecular weight will have the higher boiling point and it more likely to be solid at room temperature.
Unsaturated fat is more healthy than saturated fat and it is recommended to reduce saturated fat consumption as it was linked to many diseases.

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How many grams are in 1 mole of Ar?

Ilia_Sergeevich [38]

Answer: 39.948 grams

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Ar, or 39.948 grams

3 0

3 years ago

When an excited electron in an atom moves to the ground state, the electron

kodGreya [7K]

Answer is: (4) emits energy as it moves to a lower energy state.

Atom emits a characteristic set of discrete wavelengths, according to its electronic energy levels.

Emission spectrum of a chemical element is the spectrum of frequencies emitted due to an atom making a transition from a high energy state to a lower energy state.

Each transition has a specific energy difference.

Each element's emission spectrum is unique.

7 0

3 years ago

Read 2 more answers

How many moles of oxygen would be needed to produce 84 moles of sulfur trioxide according to the following balanced chemical equ

olganol [36]

Answer:

126 moles

Explanation:

2S +3 o2=2so3

So if 2 moles of so3 required 3 moles of oxygen

. So 84 moles of so3 will require 84*3/2=126 moles of oxygen

4 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

Approximately how much of the world’s oil and natural gas reserves are believed to be in the arctic?

Angelina_Jolie [31]

In 2009, the US Geological Survey estimated the Arctic may be home to 30% of the planet's natural gas reserves and 13% of oil.

7 0

3 years ago