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3 years ago

5mL of Ethanol has a mass of 3.9g and 5mL of benzene has a mass 4.4g. Which liquid has a higher density?

1 answer:

6 0

For Ethanol:

D = m / V

D = 3.9 g / 5 mL

D = 0.78 g/mL

For benzene:

D = 4.4 g / 5 mL

D = 0.88 g/mL

benzene has a higher density.

hope this helps!

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8.1 moles of HCl is put into a beaker with enough water to make a total of 189.98 milliliters of

Kay [80]

Answer:

42.64 M

Explanation:

Step 1: Given data

Moles of HCl (solute): 8.1 mol

Volume of the solution: 189.98 mL

Step 2: Convert the volume of the solution to liters

We will use the relationship 1 L = 1,000 mL.

$189.98 mL \times \frac{1L}{1,000mL} = 0.18998L$

Step 3: Calculate the molarity of the solution

The molarity is equal to the moles of solute divided by the liters of solution.

$M= \frac{8.1mol}{0.18998L} = 42.64 M$

8 0

3 years ago

Write a balanced molecular equation describing each of the following chemical reactions:

zhannawk [14.2K]

Answer: CaCO3 = CaO + CO2

Explanation:

CaCO3 = CaO + CO2

This is balanced already.

Elem Reactant Products

Ca 1 1

C 1 1

O 3 3

4 0

3 years ago

For each solution, calculate the initial and final pH after the addition of 0.010 mol of NaOH.

Murljashka [212]

Answer:

A. 12.6

B. 4.05

C. 10.93

Explanation:

A. 0.010mol of NaOH in 250.0mL gives a concentration of 0.04M NaOH = 0.04M OH⁻

pOH = -log [OH⁻] = 1.398

pH = 14-pOH = 12.6

B. The reaction of NaOH with HCHO₂ is:

NaOH + HCHO₂ → H₂O + CHO₂⁻ + Na⁺

Initial moles of CHO₂⁻ and HCHO₂ are:

CHO₂⁻ = 0.250L × (0.275mol/L) = 0.06875moles

HCHO₂ = 0.250L × (0.195mol/L) = 0.04875moles

After reaction:

CHO₂⁻ = 0.06875moles + 0.010mol = 0.07875mol

HCHO₂ = 0.04875moles - 0.010mol = 0.03875mol

Using H-H equation (pKa of this buffer: 3.74)

pH = 3.74 + log [CHO₂⁻] / [HCHO₂]

pH = 3.74 + log [0.07875mol] / [0.03875mol]

pH = 4.05

C. The reaction of NaOH with CH₃CH₂NH₃Cl is:

NaOH + CH₃CH₂NH₃Cl → H₂O + CH₃CH₂NH₂ + Cl⁻ + Na⁺

Initial moles of CH₃CH₂NH₃Cl and CH₃CH₂NH₂ are:

CH₃CH₂NH₃Cl = 0.250L × (0.235mol/L) = 0.05875moles

CH₃CH₂NH₂ = 0.250L × (0.255mol/L) = 0.06375moles

After reaction:

CH₃CH₂NH₃Cl = 0.05875moles - 0.010mol = 0.04875mol

CH₃CH₂NH₂ = 0.06375moles + 0.010mol = 0.07375mol

Using H-H equation (pKa of this buffer: 10.75)

pH = 10.75 + log [CH₃CH₂NH₂] / [CH₃CH₂NH₃Cl]

pH = 10.75 + log [0.07375mol] / [0.04875mol]

pH = 10.93

8 0

3 years ago

Calculate the change in entropy of 3 moles of liquid water if you heat it from 5˚C to 95˚C. The molar heat capacity of liquid wa

IceJOKER [234]

Answer: The entropy change of the liquid water is 63.4 J/K

Explanation:

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=n\times C_{p}\times \ln (\frac{T_2}{T_1})$

where,

$\Delta S$ = Entropy change

$C_{p}$ = molar heat capacity of liquid water = 75.38 J/mol.K

n = number of moles of liquid water = 3 moles

$T_2$ = final temperature = $95^oC=[95+273]K=368K$

$T_1$ = initial temperature = $5^oC=[5+273]K=278K$

Putting values in above equation, we get:

$\Delta S=3mol\times 75.38J/mol.K\times \ln (\frac{368}{278})\\\\\Delta S=63.4J/K$

Hence, the entropy change of the liquid water is 63.4 J/K

4 0

3 years ago

How does the polarity of water give it this solvent property

mafiozo [28]

Answer:

being polar, it can easily dissolve other polar substances or substances with ionic bonds like nacl

5 0

3 years ago

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