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4 years ago

Are oil and water a mixture or solution

Chemistry

2 answers:

tatyana61 [14]4 years ago

8 0

A solution because they stay as separate layers and do not form a mixture.

Dvinal [7]4 years ago

3 0

Yes oil and water is a solution not a mixture

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What volume. In liters, of H2O(g) measured at STP is produced by the combustion of 15.63 g of natural gas (CH4) according to the

fomenos

Answer:

V = 43.95 L

Explanation:

Given data:

Mass of CH₄ decomposed = 15.63 g

Volume of H₂O produced at STP = ?

Solution:

Chemical equation:

CH₄ + 2O₂ → 2H₂O + CO₂

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 15.63 g/ 16 g/mol

Number of moles = 0.98 mol

Now we will compare the moles of H₂O with CH₄.

CH₄ : H₂O

1 : 2

0.98 : 2×0.98 = 1.96 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K

V = 43.95atm.L / 1atm

V = 43.95 L

3 0

3 years ago

A gas sample is held at constant temperature. What happens to the pressure of the sample if the volume is doubled and the number

Iteru [2.4K]

Answer:

If the volume is doubled and the number of molecules is doubled, pressure is unchanged

Explanation:

Step 1: Data given

Temperature = constant

Volume will be doubled

Number of molecules will be doubles

Step 2:

p*V = n*R*T

⇒ gas constant and temperature are constant

Initial pressure = n*R*T / V

Initial pressure = 2*R*T/2

Initial pressure = RT

Final pressure = 4*RT / 4

Final pressure = R*T

If the volume is doubled and the number of molecules is doubled, pressure is unchanged

3 0

3 years ago

How many atoms are in .800 g of Ca

Goshia [24]

The answer for this question is 0.8

8 0

4 years ago

Read 2 more answers

12. Which atoms or group of atoms make up the functional group of ethanol (CH3CH2OH)?

Montano1993 [528]

Answer:

Option A. the hydroxyl group (-OH)

Explanation:

Ethanol, CH₃CH₂OH belongs to the class of organic compound called alkanol.

They have general formula as R–OH

Where

R => is an alkyl group

OH => is the hydroxyl group

The hydroxyl group (OH) is the functional group of the alkanol (alcohol)

8 0

3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago