All categories

3 years ago

What is the common name of this?

Chemistry

1 answer:

drek231 [11]3 years ago

8 0

Answer:

butyl alcohol

Explanation:

You might be interested in

What is the ph of a solution of 0.450 m kh2po4, potassium dihydrogen phosphate?

iVinArrow [24]

PKa = -log (Ka) = log [HPO4(2-)] - log[H+]^2 = - log(4.2×10^-13)
pH = - log [H+]
- log [H+]^2 = - 2 log [H+]
2pH = - log (4.2×10^-13) - log [HPO4(2-)]
2pH = - log (4.2×10^-13) - log (0.550)
pH = 6.32

8 0

3 years ago

PLZ ASAPThe model shows how the Sun fuses two nuclei of hellum-3 into beryllium-6. Since beryllium-6 is unstable, it will decay

just olya [345]

Answer:

Helium-4

Explanation:

3 0

3 years ago

28)

White raven [17]

Answer:

D [polymers]

Explanation:

The joining of monomers (small molecules) is polymerization.

6 0

4 years ago

Read 2 more answers

The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is urea. If 13.

Trava [24]

Answer:

Molality = 1.46 molal

The freezing point of the solution = -2.72 °C

Explanation:

Step 1: Data given

The freezing point of water H2O is 0.00°C at 1 atmosphere

urea = nonelectrolyte = van't Hoff factor = 1

Mass urea = 13.40 grams

Molar mass urea = 60.1 g/mol

Mass of water = 153.2 grams

Molar mass H2O = 18.02 g/mol

Kf = 1.86 °C/m

Step 2: Calculate moles urea

Moles urea = mass urea /molar mass urea

Moles urea = 13.40 grams / 60.1 g/mol

Moles urea = 0.223 moles

Step 3: Calculate the molality

Molality = moles urea / mass water

Molality = 0.223 moles / 0.1532 kg

Molality = 1.46 molal

Step 4: Calculate the freezing point of the solution

ΔT = i * Kf * m

ΔT = 1* 1.86 °C/m * 1.46 m

ΔT = 2.72 °C

The freezing point = -2.72 °C

3 0

3 years ago

It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2

Lapatulllka [165]

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

$n_{HCl}=2*n_{Mg(OH)_2}$

And in terms of volumes and concentrations we verify:

$V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}$

So we use the given data to proof it:

$4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594$

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

Best regards.

4 0

3 years ago