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3 years ago

Oil and water a combined to make salad dressing.

Chemistry

2 answers:

FinnZ [79.3K]3 years ago

8 0

Answer:

physical

Explanation:

maks197457 [2]3 years ago

5 0

When you combine any thing together that you cant take apart is a Physical Change.

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Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th

Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.

The mass percent of lithium hydroxide can be calculated with the following equation:

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$ (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$ (2)

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$

$0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}$

$\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol$

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$ (3)

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$

$\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol$

Solving for $m_{LiOH}$ , we have:

$m_{LiOH} = 0.082 g$

Hence, the percent lithium hydroxide is (eq 1):

$\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%$

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

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I hope it helps you!

5 0

2 years ago

What is the difference between the product and the reactant

timama [110]

Answer:

Explanation:

A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction.

h2 + 02 = h2o

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8 0

3 years ago

Which of the following statement is FALSE? a. Lowering the free energy of the transition state can increase a reaction rate. b.

ryzh [129]

Answer: Option (c) is the correct answer.

Explanation:

Activation energy or free energy of a transition state is defined as the minimum amount of energy required to by reactant molecules to undergo a chemical reaction.

So, when activation energy is decreased then molecules with lesser amount of energy can also participate in the reaction. This leads to an increase in rate of reaction.

Also, increase in temperature will help in increasing the rate of reaction.

Whereas at a given temperature, every molecule will have different energy because every molecule travels at different speed.

Hence, we can conclude that out of the given options false statement is that at a given temperature and time all molecules in a solution or a sample will have the same energy.

8 0

3 years ago

an unknown amount of mercury (ii) oxide was decomposed in the lab. mercury metal was formed and 4.50 l of oxygen gas was release

Assoli18 [71]

The initial weight of mercury oxide in the sample was 59.1 g HgO.

<h3>Steps</h3>

chemical reaction

2HgO ⟶ 2Hg + O₂

the moles of O₂

pV = nRT

n = (pV)/(RT)

Data:

p = 0.970 atm

V = 4.50 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 390.0 K

Calculation:

n = (0.970 × 4.500)/(0.082 06 × 390.0)

n = 0.1364 mol O₂

the moles of HgO

The molar ratio is 1 mol O₂/2 mol HgO.

Moles of HgO = 0.1364 mol O₂ × (2 mol Hg/1 mol O₂)

Moles of HgO = 0.2728 mol HgO

the mass of HgO

Mass of HgO = 0.2728 mol HgO × (216.59 g HgO/1 mol HgO)

Mass of HgO = 59.1 g HgO

For the creation of various organic mercury compounds and specific inorganic mercury salts, mercury(II) oxide, or HgO, serves as a source of elemental mercury.

This red or yellow crystalline substance is also utilised in mercury batteries and zinc-mercuric oxide electric cells as an electrode (combined with graphite).

<h3>What is the purpose of mercury oxide?</h3>

Mercuric oxide is a colourless, crystalline powder that ranges from yellow to orange-yellow.

It serves as a seed protectant, a pigment, a preservative, and an ingredient in alkaline batteries and cosmetics.

<h3>Is there a combination of mercury oxide?</h3>

The powder form of mercury oxide is dark black or dark brown.
An intimate blend of metallic mercury and mercuric oxide rather than a genuine compound.

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4 0

1 year ago