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3 years ago

During the synthesis of a triglyceride from its subunits, how many molecules of water are produced from dehydration synthesis?

Chemistry

1 answer:

Free_Kalibri [48]3 years ago

7 0

Answer:

5 synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis

Explanation:

synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis synthesis

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Find the mass in kilograms of the liquid air that is required to produce 600L of oxygen. In normal condition, 1L of liquid air h

IgorC [24]

Mass in kilograms of liquid air required = 0.78 kg

Given that

1 Litre of liquid air contains 1.3 grams of oxygen ( air )

Determine the amount of liquid air in Kg

volume of air given = 600 L

mass of liquid air required = x

1 litre = 1.3 grams

600 L = x

∴ x ( mass of liquid air ) = 1.3 * 600

= 780 g = 0.78 kg

Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg

Learn more about liquid air : brainly.com/question/636295

3 0

2 years ago

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

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3 years ago

Hammerhead Shark Process of Sexual Reproduction

ICE Princess25 [194]

I don’t even know that’s just weird

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3 years ago

What are oil and waters properties of matter?

yulyashka [42]

Answer:

Two other beneficial properties of the oil include the inhibition of corrosion of metal surfaces, such as steel, and the removal of dirt and detritus via detergency.

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2 years ago

Calculate the number of grams in magnesium present in 25 g of magnesium phosphate

olga55 [171]

you can google it and it pops up right away

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3 years ago

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