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3 years ago

_______are lipids that contain cholesterol

Chemistry

1 answer:

alukav5142 [94]3 years ago

3 0

Answer: Fats

Explanation: Cholesterol is one of several types of fats (lipids) that play an important role in your body.

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When 412.5g of calcium carbonate react with 521.9g of aluminum fluoride,how many grams of each product can be produced

aliya0001 [1]

<h3><u>Answer;</u></h3>

321.8 g CaF2

321.5 g Al2(CO3)3

<h3><u>Explanation;</u></h3>

The equation for the reaction is;

3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2(CO3)3

Number of moles of CaCO3 will be;

=(412.5 g CaCO3) / (100.0875 g CaCO3/mol)

= 4.12139 mol CaCO3

Number of moles of AlF3 will be;

= (521.9 g AlF3) / ( 83.9767 g AlF3/mol)

= 6.21482 mol AlF3

But;

4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.

Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.

Therefore;

Mass of CaF2 will be;

(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2

Mass of Al2(CO3)3 on the other hand will be;

(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2(CO3)3/mol) = 321.5 g Al2(CO3)3

3 0

4 years ago

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha

lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

$\frac{60}{100}\times 3.6 = 2.16 L$

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator =50% of 3.6 L = $\frac{50}{100}\times 3.6=1.8 L$

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired volume

2.16 L - 60% of( x) = 1.8 L

$2.16 L-0.6x=1.8 L$

$x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L$

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0

4 years ago

Which of the following species has bonds with

Rzqust [24]

The species bonds with the most ionic character is the compound with elements that develop the higher electronegativity difference between each other.
the most ionic character had maximum EN.
EN is max for SiO2

hope this help

4 0

3 years ago

Using your value of Ksp, and starting with an equilibrium system consisting of a saturated solution of calcium hydroxide, predic

Ugo [173]

Answer:

(A) $HNO_{3}$ Solubility will increase.

(B) $NaOH$ Solubility will decrease.

(C) $Ca(NO_{3} )_{2}$ Solubility will decrease.

Explanation:

a) According to Le Chatelier's Principle when a change is introduced in a reaction system at equilibrium, the system responds by the reaction shifting in the direction to counter the change introduced.

When the change introduced is addition of acid, some $OH^{-}$ are consumed due to it and the system would respond by reaction shifting towards creation of more $OH^{-}$ in the system by increased dissolution. Thus solubility increase.

(b) When $OH^{-}$ are added, the reaction shifts to counter the change introduced of increased $OH^{-}$ by shifting to side that decreases the increased $OH^{-}$ . That happens by reaction shifting towards some $Ca (OH)_{2}$ precipitating back. Hence solubility decreases.

(c) The change introduced is addition of $Ca^{2+}$ which will be countered by reaction shifting to side that decreases increased $Ca ^{2+}$ , which happens by some $Ca(OH)_{2}$ precipitating back. Hence solubility decreases.

6 0

4 years ago

At STP, what is the volume of 55.0 g of BF3?

worty [1.4K]

Answer:

The volume of 55 grams of BF₃ is 18.1664 L.

Explanation:

If the molar mass of compound BF₃ is 67.81 $\frac{grams}{mole}$ , then the number of moles 55 grams of compound contains is calculated as:

$55 grams* \frac{1 mole}{67.81 grams} = 0.811 moles$

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then it is possible to apply the following rule of three: if by definition of STP 1 mole of a compound occupies 22.4 L, 0.811 moles of the compound occupies how much volume?

$volume=\frac{0.811 moles* 22.4 L}{1 mole}$

volume= 18.1664 L

<u><em>The volume of 55 grams of BF₃ is 18.1664 L.</em></u>

8 0

3 years ago