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4 years ago

What is formed when amino acids form long chains?

1 answer:

4 0

Amino acids form long chains/chains in general with peptide bonds so they make polypeptides that then turn into proteins

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Buckets of paint can be classified as what type of mixture?

marysya [2.9K]

Answer:

Homogeneous

Explanation:

Homogeneous mixtures are uniform in composition. They have the same proportion of components throughout. Homogeneous mixtures are called solutions. Sugar, paint, alcohol, gold are all examples of homogeneous mixtures because they look the same throughout.

6 0

3 years ago

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

3 years ago

What is the salt produced when sulfuric acid (H2SO4) reacts with potassium hydroxide (KOH)?

pashok25 [27]

I got B. Because the answer is K2SO4, I got it right on my test so I know its right

8 0

3 years ago

Read 2 more answers

A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),

Semenov [28]

Answer:

[KOH] : 1.47 M

[KOH] : 1.22 m

[KOH]: 6.42 % mass percent.

Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

7 0

3 years ago

Identify if the following are chemical or physical properties: "CP" for

mel-nik [20]

Answer:

1.pp

2.cp

3.pp

4.pp

5.pp

6.pp

8 0

4 years ago