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3 years ago

7

Identify the following as an example of a physical or chemical. Olive oil, vinegar, salt, and pepper are shaken together togethe

r to make salad dressing

1 answer:

erastova [34]3 years ago

3 0

Answer:

Physical

Explanation:

Olive oil, vinegar, salt, and pepper are shaken together to make salad dressing. Physical or chemical change?

Physical

You might be interested in

The degree to which various compounds will dissociate in solution varies greatly.

uranmaximum [27]

Answer: True

Explanation:

Weak electrolytes are those solutions which do not undergo complete dissociation when dissolved in water. The dissociation of weak electrolytes is given by an equilibrium.

Example: $CH_3COOH\rightleftharpoons CH_3COO^-+H^+$

Strong electrolytes are those solutions which undergo complete dissociation when dissolved in water. The dissociation of strong electrolytes is given by a right arrow.

Example: $HCl\rightarrow H^++Cl^-$

Thus the degree to which various compounds will dissociate in solution varies greatly is true.

5 0

3 years ago

In which way would a drought affect how rodents get their energy?

love history [14]

Answer:

<h3>When a drought occurs, their food supply can shrink and their habitat can be damaged. ... Losses or destruction of fish and wildlife habitat. Lack of food and drinking water for wild animals. Increase in disease in wild animals, because of reduced food and water supplies.</h3><h3>While insects and cacti might provide a meagre supply of water, most desert animals survive by being what Price calls "water misers". ... To perform this feat, they have evolved specialized kidneys with extra microscopic tubules for extracting water from urine.</h3>

Explanation:

3 0

2 years ago

A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.

makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4' to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

% => grams => moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H => 7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio, divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0

3 years ago

Dry air is 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.039% carbon dioxide so varying amounts of water vapor- depending on hu

AfilCa [17]

Answer:

C. Carbon dioxide

Explanation:

Carbon dioxide is one of the end-product of combustion reactions involving many fuels today.

With the rapid increase in urbanization and technological development, man demand for energy increased tremendously. The discovery of fossil fuels paved the way for the astronomical increase in the concentration of carbon dioxide in the atmosphere. The burning of fossil fuels like coal and oil invovles the process where the carbon atoms present in these fuels combine with oxygen in the air to make CO2. This has resulted in an increase in the concentration of atmospheric carbon dioxide (CO2).

The burning fossil fuels for electricity, industry, heat, and transportation are the major sources of the emossion of carbon dioxide.

Also, the cutting down of trees for paper production, building construction and for the establishment of settlements also increase the concentration of carbon dioxide in the atmosphere. Trees are help remove carbon dioxide from the atmosphere through the process of photosynthesis. However, when these trees are cut down, carbon dioxide accumulates in the atmosphere.

5 0

3 years ago

A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa

Rudik [331]

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

<em>1 mole of HCl reacts per mole of ammonia.</em>

Mass of NH3 is obtained as follows:

<em>Moles HCl:</em>

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>

<em>Mass NH3 in the aliquot:</em>

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

<h3>1.43 (w/w %)</h3>

6 0

3 years ago