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3 years ago

What chemicals need to make liquid bakelite ?

1 answer:

5 0

It begins with heating of phenol and formaldehyde in the presence of a catalyst such as hydrochloric acid, zinc chloride, or the base ammonia. This creates a liquid condensation product, referred to as Bakelite A, which is soluble in alcohol, acetone, or additional phenol.

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At a certain temperature the speeds of six gaseous molecules in a container are 2.0 m/s, 2.2 m/s, 2.6 m/s, 2.7 m/s, 3.3 m/s, and

iren [92.7K]

Answer:

root-mean-sqaure = 2.77 m/s

average = 2.72 m/s

The root-mean-square is always the largest because it takes account of the variance of the spread of the data. The increase is related to the fact that the data varies to sample.

Explanation:

The rootmean-square (R) is the square root of the squares of the valeus divided by the number of the datas.

$R = \sqrt{\frac{x1^2 + x2^2 +...+xn^2}{n} }$

$R = \sqrt{\frac{(2.0)^2 + (2.2)^2 + (2.6)^2 + (2.7)^2 ^(3.3)^2 + (3.5)^2}{6} }$

R = √(46.03)/6

R = 2.77 m/s

The average speed is the sum of the speeds divided by the number of datas:

$A = \frac{2.0 + 2.2 + 2.6 + 2.7 + 3.3 + 3.5}{6}$

A = 16.3/6

A = 2.72 m/s

5 0

3 years ago

Read 2 more answers

At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of

Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.031M/atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

$C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = $0.0905mol/L$

Volume of solution = 425 mL

Putting values in above equation, we get:

$0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g$

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

8 0

3 years ago

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

5 0

4 years ago

Pb(NO3)2 + K2CO3 --> PbCO3 + KNO3 A. 5 B. 2 C. 4

emmasim [6.3K]

Answer:

Pb(NO3)2 + K2CO3 ----> PbCO3 + 2KNO3

The answer is B. 2

Hope this helps!

4 0

4 years ago

Blue and orange, which color can be best absorbed?

baherus [9]

Blue is the color that can be best absorbed, as orange is the color seen.

6 0

3 years ago