All categories

3 years ago

What type of reaction occurs when complex compounds are broken down into simpler substances?

1 answer:

4 0

Complex compounds are broken down to simpler substances in catabolic reactions.

These kinds of reactions often occur in biological systems. In living organisms, complex compounds like lipids, proteins and complex sugar like cellulose are broken down into simpler forms. Products of these reactions are simple sugars, amino acids etc. but a certain amount of energy is also produced and stored in energy molecules for future use.

You might be interested in

How is the density of a substance calculated?

murzikaleks [220]

Example:

sample density of gasoline, 20 g of weigth into 5 mL

Answer:

D = m / V

D = 20 g / 5 mL

D = 4 g/mL

3 0

3 years ago

If I initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 K, and then I raise the pressure

CaHeK987 [17]

Answer : The volume of gas will be 29.6 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 12 atm

$P_2$ = final pressure of gas = 14 atm

$V_1$ = initial volume of gas = 23 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 200K

$T_2$ = final temperature of gas = 300K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{12atm\times 23L}{200K}=\frac{14\times V_2}{300K}$

$V_2=29.6L$

Therefore, the new volume of gas will be 29.6 L

5 0

2 years ago

Read 2 more answers

CO2 + NaOH <=> NaHCO3 Balance

NeTakaya

Answer:

Explanation:

It is balaned

Left side

1 Na

1 C

3 Os

1 H

=========

Right side

1 Na

1 H

1 C

3 Os

5 0

3 years ago

Read 2 more answers

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

A helium-filled balloon has a volume of 2.48 L and a pressure of 150 kPa. The volume of the balloon increases to 2.98 L after yo

Tema [17]

Answer:

THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.

AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.

Explanation:

The question above follows Boyle's law of the gas law as the temperature is kept constant.

Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Mathematically, P1 V1 = P2 V2

P1 = 150 kPa = 150 *10^3 Pa

V1 = 2.48 L

V2 = 2.98 L

P2 = ?

Rearranging the equation, we obtain;

P2 = P1 V1 / V2

P2 = 150 kPa * 2.48 / 2.98

P2 = 372 *10 ^3 / 2.98

P2 = 124.8 kPa.

The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.

6 0

3 years ago