All categories

3 years ago

What did sewage treatment in Ancient Rome and 19th century London have in common?

Chemistry

1 answer:

dolphi86 [110]3 years ago

5 0

Answer:

The history of water supply and sanitation is one of a logistical challenge to provide clean water and sanitation systems since the dawn of civilization. Where water resources, infrastructure or sanitation systems were insufficient, diseases spread and people fell sick or died prematurely.

Explanation:

You might be interested in

A bottle of wine contains 12.8% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm. Calculate the concentratic etha

Scrat [10]

Answer : The mass percent of ethanol is, 10.38 % and the molality of ethanol is, 2.52 mole/kg

Explanation :

In wine, the solute and solvent are ethanol and water respectively.

Given :

12.8 % ethanol by volume means 12.8 mL ethanol present in 100 mL solution.

Volume of ethanol = 12.8 mL

Volume of solution = 100 mL

Volume of water = 100 - 12.8 = 87.2 mL

Density of ethanol = $0.789g/cm^3=0.789g/mL$

Density of water = 1 g/mL

Now we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 12.8mL=10.1g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/mL\times 87.2mL=87.2g$

Now we have to calculate the total mass of 100 mL of wine.

Total mass of 100 mL of wine = 10.1 + 87.2 = 97.3 g

Now we have to calculate the mass percent of ethanol.

$\text{Mass of percent of ethanol}=\frac{\text{Mass of ethanol}}{\text{Total mass of solution}}\times 100$

$\text{Mass of percent of ethanol}=\frac{10.1g}{97.3g}\times 100=10.38\%$

The mass percent of ethanol is, 10.38 %

Now we have to calculate the molality.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Molar mass of ethanol = 46 g/mole

$Molality=\frac{10.1g\times 1000}{46g/mole\times 87.2g}=2.52mole/kg$

The molality of ethanol is, 2.52 mole/kg

6 0

4 years ago

If 0.400 moles CO and 0.400 moles O2 completely react how many grams of CO2 are produces?

True [87]

If 0.400 moles CO and 0.400 moles O2 completely react, 17.604 grams of CO2 would be produced.

First, let us look at the balanced equation of reaction:

$2CO + O_2 ---> 2CO_2$

According to the equation, the mole ratio of CO and O2 is 2:1. But in reality, the mole ratio supplied is 1:1. Thus, CO is the limiting reactant while O2 is in excess.

Also from the equation, the ratio of CO consumed to that of CO2 produced is 1:1. Thus, 0.400 moles of CO2 would also be produced from 0.400 moles of CO.

Recall that: mole = mass/molar mass

Therefore, the mass in grams of CO2 that would be produced from 0.400 moles can be calculated as:

Mass = mole x molar mass

= 0.400 x 44.01

= 17.604 grams

More on calculating mass from number of moles can be found here: brainly.com/question/12513822

7 0

3 years ago

Organic molecules are defined as chemical compounds that contain ______ in distinct ratios and structures.

OLga [1]

Answer:

The answer is: <em>carbon</em>

Explanation:

Organic molecules contain the chemical element carbon (C) in its structure. In this type of molecules, carbon is usually bonded to hydrogen (H), oxygen (O) and, with less frecuency, nitrogen (N). Therefore, in these molecules, carbon forms simple, double and triple bonds with itself. Examples of organic molecules that are very important in biology are carbohydrates, lipids, proteins and nucleic acids.

8 0

4 years ago

How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ

Jet001 [13]

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

r = 7.12×10⁻³ m

r = 7.12 mm

Hence the distance between the point charge = 7.12 mm

3 0

3 years ago

An isotope of cesium-137 has a half-life of 30 years. If 5.0g of cesium-137 decays over 60 years. How many grams will remain.

LenKa [72]

Answer:

1.25 gram of cesium-137 will remain.

Explanation:

Given data:

Half life of cesium-137 = 30 year

Mass of cesium-137 = 5.0 g

Mass remain after 60 years = ?

Solution:

Number of half lives passed = Time elapsed / half life

Number of half lives passed = 60 year / 30 year

Number of half lives passed = 2

At time zero = 5.0 g

At first half life = 5.0 g/2 = 2.5 g

At 2nd half life = 2.5 g/ 2 = 1.25 g

Thus. 1.25 gram of cesium-137 will remain.

6 0

3 years ago