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2 years ago

What natural resource did the Mesopotamians use to protect their cities from floods?

Chemistry

2 answers:

Zielflug [23.3K]2 years ago

7 0

Answer:

mud bricks.

Explanation:

pickupchik [31]2 years ago

5 0

Answer: Mud Bricks

Explanation: Fertile Crescent has important natural resources that encourage settlement including Mud

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A chemical bond between two atoms results from a simultaneous

Elena L [17]

If you are asking for the word for this definition it is a attraction by the two nucluei

7 0

2 years ago

Read 2 more answers

How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)? 2AgNO3 + CaC

nexus9112 [7]

Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction

2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).

From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂

Henceforth, 6.2 moles of AgNO₃ reacts with $\frac{6.2}{2}$ = 3.1 moles of CaCl₂.

1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-

A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.

C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.

D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃

Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.

6 0

2 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

$m=\frac{92.5}{100}\times 0.5940=0.54945 g$

1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

6 0

2 years ago

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

Answer: The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0

2 years ago